answers (partially)

100)
i) straight lines through 0.
ii) x_u = (1, v, v²) and x_v = (0, u, 2uv) are dependent if u=0. But this point is singular.
iii) The normal vector has the same or opposite direction as x_u ⊗ x_v, or, equivalently, (v², -2v, 1). Tangent plane v²x₁ - 2v x₂ + x₃ = 0.
iv) The equation of the tangent plane is independent of u.

110) The first fundamental form is 5(du)² + u²(dv)², so a_{1 1} = 5, a_{1 2} = 0, a_{2 2} = u².
From cos(φ) = (Σ a_{i j} da_i db_j)/(√(Σ a_{i j} da_i da_j)√(Σ a_{i j} db_i db_j)) with a₁ = b₂ = t² en a₂ = b₁ = t³, follows, if t=1, cos(φ) = 36/(7√29).

111) x_u = (1,2u,v) and x_v = (1,2v,u), so a_{1 2} = 0 if uv = -1/5, so x = (v - 1/(5v), v² + 1/(25v²), -1/5), so y - x² = 2/5 and z = -1/5.

112) (ds)² = (dθ)² + sin²(θ) (dφ)².
The surface element is then √ sin²(θ) dθ dφ.
So the required surface is ₀ ∫ ^2π _θ2 ∫ ^θ1 sin(θ) dθ dφ = 2π (cos(θ2) - cos(θ1)).

113) Take the following parametrisation of the plane: (u√2) (cos(v/√2), sin(v/√2), 0). Then both surfaces have first fundamental form 2(du)² + u² (dv)².

117) With φ(u,v) = u we find φ_u = 1, φ_v = 0 ; so the differential equation becomes:
du/dv = (a_{2 2}φ_u - a_{1 2}φ_v)/(a_{1 1}φ_v - a_{1 2}φ_u) = -a_{2 2}/a_{1 2} = -(2+u²)/(uv).
We get u du/(2+u²) = -dv/v, so ln(2+u²) = -2 ln |v| + c. Hence it follows that v²(2+u²) = μ.

118) The meridians are determined by φ(θ,φ) = λ, with φ(θ,φ) = φ, so by φ = λ.
We get φ_θ = 0 and φ_φ = 1, so the differential equation becomes: cos(α) = (-a_{1 1}dθ - a_{1 2}dφ)/(√a_{1 1}*√(a_{1 1}dθ² + 2 a_{1 2}dθ dφ + a_{2 2}dφ²).
With a_{1 1} = 1, a_{1 2} = 0 and a_{2 2} = sin²(θ) we find cos(α) = -dθ / √(dθ² + sin²(θ) dφ²), so dθ/sin(θ) = +tg(α) dφ.
Then we get for the loxodromes: φ + c = + (1/2) cotg(α) ln(|(1+cos(θ))/(1-cos(θ))|).

127) f ' = 0, so f is constant; then we have a right helicoid surface. Or h=0, surface of revolution.

128) (ds)² = (du)² + (u²+h²) (dv)² , so a_{1 1} = 1, a_{1 2} = 0, a_{2 2} = u²+h².
φ(u,v) = u, hence φ_u=1, φ_v=0. We get in 116:
(1/2)√2 = (u²+h²) dv / (√(u²+h²) √((du)² + (u²+h²)(dv)²)), dus: +du/√(u²+h²) = dv.
Hence we find: ln | u + √(u²+h²) | = +v + c.

129) The parameter lines are perpendicular to each other in the intersection points where uv = b²-a².

130) Let x(u₁(t),u₂(t)) be such a bissectrix.
By equating the angle with the u₁-line and the angle with the u₂-line, we find via 116 (with φ(u₁,u₂) = u₁ and φ(u₁,u₂) = u₂ respectively):
(a_{1 2} du₁ + a_{2 2} du₂)/√a_{2 2} = +(a_{1 1} du₁ + a_{1 2} du₂)/√a_{1 1}.
Hence we find (provided that a_{1 2}² - a_{1 1} a_{2 2} ≠ 0) : a_{1 1} (du)² = a_{2 2} (dv)².

131) Apply 120.

137) The parameter lines form an orthogonal net because a_{1 2} = 0.
The equation of the indicatrix is du dv = 1 (unless b=0), because h_{1 1} = h_{2 2} = 0 and h_{1 2} = -b/√(b²+(a+u)²) (check this). This is an orthogonal hyperbola with asymptotes du = 0 and dv = 0.
The net of the asymptotic lines is du dv = 0, this is the net of the parameter lines.

145) Check by calculation that a_{1 1} = 1 + (f ' )², a_{1 2} = 0, a_{2 2} = u²; h_{1 1} = f " / √(1 + (f ' )²), h_{1 2} = 0, h_{2 2} = u f ' / √(1 + (f ' )²).
With h_{1 1}a_{2 2} - 2 h_{1 2}a_{1 2} + h_{2 2}a_{1 1} = 0 we get f ' / (1 + (f ' )²) = -uf ".
Consider this as a differential equation in f ' (let f ' = g), and find after separating the variables with partial fraction decomposition (f ')² = c²/(u²-c²).
We find f = +c log((u + √(u²-c²)/c), so u = c cosh(f/c).

The catenary is the curve formed by a homogeneous cord under the influence of gravity. It looks like a parabola, but is essentially different.
The form of the catenoid is visible when we span a soap fleece between two circular thread figures whose centres are lying straight above each other.

Extra problem :

a) (see picture above)
u-lines: parabolas in planes z=constant, x≥0.
v-lines: parabolas in planes x=constant, z≥0.

b) First fundamental form (4u²+v²)(du)² + 2uv du dv + (u²+4v²)(du)².
The parameter lines are perpendicular where uv=0, so on the x- and z-axis.

c) Second fundamental form (2/√((u²+v²)²+2u²v²) (v²(du)² - 2uv du dv + u²(dv)²).
Indicatrix (2/√..) (v du - u dv)² = 1, two parallel lines with direction vector x_u + (v/u)x_v.
Asymptotic lines du/dv = u/v, u=cv, x = v²(c²,c,1) (halfs of straight lines).

d) Orthogonal trajectories of the u-lines from x_u . (x_u+ x_v dv/du) = 0.
We get du/dv = - (4u²+v²)/(uv) and at the end v²u² + 4u⁴ = constant.

e,f) From c) we see the surface exists of half straight lines v²(c²,c,1) = λ(t²,t,1) (λ positive). The half lines start from (0,0,0) (top) and go through (t²,t,1) (direction curve, a parabola). So it is a half parabolic cone.
Tangent planes (equation v²x - 2uv y + u²z = 0) are constant on rules u = cv (equation x - 2cy + c²z = 0).

f) Parametrisation λ(t²,t,1): see e) above.
The rules are the λ-lines, so the orthogonal trajectories of the rules satisfy x_λ . (x_λ+ x_t dt/dλ) = 0. We get dt/dλ = -(t⁴+t²+1)/((2t³+t)λ).
At the end we find λ||(t²,t,1)|| = constant, so an orthogonal trajectory exists of points at the same distance from (0,0,0).

153) If the meridian curves of a helicoid surface are curvature lines, then dv = 0 must be a solution of 148, so a_{1 1} h_{1 2} - a_{2 1} h_{1 1} = 0.
We get h (1 + (f ' )² + u f ' f " ) = 0. So h = 0 of f ' f " /(1 + (f ' )²) = -1/u. This gives the alleged solutions.

154) Check by calculation that a_{1 2} = h_{1 2} = 0 and use proposition 152.

155) Check by calculation that a_{1 2} = h_{1 2} = 0 and use proposition 152.

165) x ' = 0 gives a cone; e ' = 0 gives a cylinder. Suppose x ' , e ' ≠ 0.
If y(u) = x(u) + v(u) e(u) is the turning curve, then y '(u) = x '(u) + v(u) e '(u) + v '(u) e(u) has the same or opposite direction as e(u) (since this holds for x ' + v e ' because that's how we chose v).
So the ruled surface is a surface of tangents.

EP) First prove that the singular points are the points with u=v, and determine the candidate turning curve x(t) with t:=u=v and the corresponding surface of tangents x(t) + λ x '(t).
Use a change of coordinates to show that this surface of tangents coincides with the original surface.

180) x = q(cos(t),sin(t),t/p) (circular helix) ; x = (t,t²,t³) (cubic parabola).

181) x = (t cos(t) - sin(t), t sin(t) + cos(t)) (evolvent of a circle).

189) Check that the Christoffel equations 186 are in this case u^.. = v^.. = 0

190) Check what the differential equations 186 are in this case. We get:
u(u ^.)² - u(v ^.)² + (1+u²) u ^.. = 2 u ^. v ^. + u v ^.. = 0.
Furthermore we have (ds)² = (1+u²)(du)² + u²(dv)².
From the second Christoffel equation we deduce v ^. = c/u², so
u⁴(dv)² = c²(ds)² = c²((1+u²)(du)² + u²(dv)²).
Hence we get v = + ∫ (c/u) √((1+u²)/(u²-c²)) du.
For c=0 we find the meridians.

191) Start from x = (u cos(v), u sin(v), a*u). We find:
(ds)² = (1+a²)(du)² + u²(dv)², -u (v ^.)² + (1+a²) u ^.. = 2 u ^. v ^. + u v ^.. = 0.
Like in 190, u⁴(dv)² = c²(ds)². Here this yields:
u⁴(dv)² = c²(1+a²)(du)² + u² c² (dv)², from which we get:
v = +√(1+a²) ∫ c/(u √(u²-c²)) du.
With t = √(u²-c²) we find v = +√(1+a²) arctan(√(u²-c²)/c) + d.

Notice that the solutions in 190 en 191 satisfy the fact that in each point a geodesic departs in each direction.

199)
a) Check by calculation that the indicatrix becomes (2/√...) (du - dv)² = 1.
Each point is parabolic, but there are no singular points (because N is nowhere 0). So the surface must be a cylinder.
We find asymptotic curves with du = dv, u = v + c, so their parametrisations are (1+c, c, c²) + v(0, 1, 1). So we have the direction curve (1+c, c, c²) and the axis has direction (0,1,1).

b) A simpler parametrisation is (1+w, w, w²) + v(0, 1, 1).
The differential equations become:
(2+4w²)(w ^.)² + 2(1+2w)(w ^.)(v ^.) + 2(v ^.)² = 1
((4w-2)/(4w²-4w+3))(w ^.)² + (w ^..) = 0
((4-4w)/(4w²-4w+3))(w ^.)² + (v ^..) = 0
Solutions are, among others: w(s) = c₁, v(s) = +(1/2)√2 s + c₂ (the rules).

200) Both surfaces have first fundamental form r² (du)² + 2 (dr)². The first surface is right circular cone, the second a plane.

201)
a) We find the asymptotic lines from (1/u)(du)² - (2/v)(du)(dv) + (u/v²)(dv)² = 0 = (du/u - dv/v)², so each point is parabolic.
Solving du/dv = u/v yields u = cv. Parameter representation of the asymptotic lines v(c², c, 1/c).
b) From a) we see the surface has parameter representation v(c², c, 1/c), so it exists of straight lines through 0. Hence it is a cone, so it's developable.

203)
a) (x_u . x_v)/(||x_u||.||x_v||) = cos(α), so a_{1 2} = cos(α) √(a_{1 1}a_{2 2}).
b) Let x(u,v(u)) be such a trajectory. Require ((x_u) . (x_u + x_v dv/du))/(||x_u||.||x_u + x_v dv/du||) = (1/2)√3, so (a_{1 1} + a_{1 2} dv/du)/(√a_{1 1} . √(a_{1 1} + 2 a_{1 2} dv/du + a_{2 2}(dv/du)²)) = (1/2)√3.

204)
a) The u-lines are congruent to the graph of g; the v-lines are straight lines; U is a 'wave-like plate' (cylinder with sinusoid as direction curve).
b) We get a simpler parametrisation of U by taking (w,u,g(u)) with first fundamental form (1+g'(u)²) (du)² + (dv)².
The plane z=0 has parametrisation (w,f(u),0) with first fundamental form f '(u)² (du)² + (dv)².
Now take f so that f '(u) = √(1+g'(u)²).

205)
a) Consider two surfaces x(u,v) and y(u,v).
If two curves on the first surface have a common tangent in a certain point, they are image curves of two curves in the u,v-plane that in the original of this contact point have a common tangent with direction coefficient dv/du. So then the common tangent to these image curves has in that contact point direction x_u + x_v dv/du.
Under the diffeomorfism yox^-1 they are mapped to two curves on the second surface that have a common tangent with direction y_u + y_v dv/du.
So having a common tangent is invariant under all diffeomorfisms, not only under isometries.
b) If we bend a right circular cylinder to a plane, the circle becomes a straight line, so curvature is not invariant.

207)
a) Using u dv + v du = 0, we find a family of asymptotic lines on the surface. We get uv = c, and hence we find the alternative parametrisation (c, c², c³) + u(c, c, c²). With this parametrisation, the u-lines are straight lines.
b) x_u ⊗ x_c = (-c³ - uc², 3c³ + uc² - c², c - 2c²) is dependent on u, so the tangent plane is not constant along the rule, so the answer is no.
Alternative : Using x_u ⊗ x_c = 0 we find two singular points, namely (0,0,0) and (1/4,0,0). A cylinder doesn't have any singular points, a cone has one singular point, a surface of tangents has a whole curve of singular points.
c) Take v=1 and u=t (or c=u=t).
d) The normal vector on the surface has the same or opposite direction as (-2t³, 4t³-t², t-2t²), the principal normal is perpendicular to x ' ⊗ x " = (24t², -12t-12t², 4).
If k is a geodesic, N and n must have the same or opposite directions in all points of k, so (-2t³, 4t³-t², t-2t²) and (24t², -12t-12t², 4) must be perpendicular for all t. This doesn't even hold for t=1. So k is no geodesic.