<s>[INST] Finding the Asymptotic Curves of a Given Surface I have to find the asymptotic curves of the surface given by $$z = a \left( \frac{x}{y} + \frac{y}{x} \right),$$ for constant $a \neq 0$.
I guess that what was meant by that statement is that surface $S$ can be locally parametrized by $$X(u,v) = \left( u, v, a \left( \frac{u}{v} + \frac{u}{v} \right) \right).$$  Do you think that my parametrization is correct (meaning that I read the description of the surface correctly), and do you know of a more convenient parametrization?
Assuming that parametrization, I derived the following ($E$, $F$, $G$, are the coefficients of the first fundamental form; $e$, $f$, $g$ are coefficients of the second fundamental form; $N$ is the normal vector to surface $S$ at a point; these quantities are all functions of local coordinates $(u,v)$):
$$E = 1 + a^2 \left( \frac{1}{v} - \frac{v}{u^2} \right)^2,$$ 
$$F = -\frac{a^2 (u^2 - v^2)^2}{u^3 v^3},$$
$$G = 1 + a^2 \left( \frac{1}{u} - \frac{u}{v^2} \right)^2.$$
$$N = \frac{1}{\sqrt{E G - F^2}} \left( a \left( \frac{v}{u^2}-\frac{1}{v} \right), a \left( \frac{u}{v^2}-\frac{1}{u} \right), 1 \right).$$
$$X_{u,u} = \left( 0,0, \frac{2 a v}{u^3} \right), X_{u,v} = \left( 0,0, -a \left( \frac{1}{u^2} + \frac{1}{v^2} \right) \right), X_{v,v} = \left( 0, 0, \frac{2 a u}{v^3} \right).$$
$$e = \frac{2 a v}{u^3 \sqrt{E G - F^2}},$$
$$f = - \frac{a (\frac{1}{u^2} + \frac{1}{v^2})}{\sqrt{E G - F^2}},$$
$$g = \frac{2 a u}{v3 \sqrt{E G - F^2}}.$$
Thus, the Gaussian curvature (from these calculations) is:
$$K = -\frac{a^2 u^4 v^4 (u^2 - v^2)^2}{(u^4 v^4 + 
  a^2 (u^2 - v^2)^2 (u^2 + v^2))^2}.$$
And the mean curvature would be:
$$H = \frac{a u^3 v^3 (u^4 + v^4)}{(u^4 v^4 + a^2 (u^2 - v^2)^2 (u^2 + v^2))^{3/2
 }}.$$
So, the principal curvatures are:
$$k_{\pm} = H \pm \sqrt{H^2 - K} = a u^2 v^2 \frac{u v (u^4 + v^4) \pm \sqrt{(u^2 + v^2) (a^2 (u^2 - v^2)^4 + u^2 v^2 (u^6 + v^6))}}{(u^4 v^4 + a^2 (u^2 - v^2)^2 (u^2 + v^2))^{3/2}}.$$
In order to find the asymptotic curves, but trying to avoid the differential equation, I was hoping to find the angles $\theta (u,v)$ such that the normal curvature would always be $0$.  In other words I was trying:
$0 = k_n = k_{+} \cos{(\theta)}^2 + k_{-} \sin{(\theta)}^2$, and solving for $\theta$.
Assuming sufficient niceness, this calculate would result in: $$(u v (u^4 + v^4) + \sqrt{(u^2 + v^2) (a^2 (u^2 - v^2)^4 + u^2 v^2 (u^6 + v^6))}) \cos{(\theta)}^2 + (u v (u^4 + v^4) - \sqrt{(u^2 + v^2) (a^2 (u^2 - v^2)^4 + u^2 v^2 (u^6 + v^6))}) \sin{(\theta)}^2$$
First of all, is this approach (solving for $\theta$ rather than solving the differential equation) valid? If it is, after I find that angle $\theta$, determined by location $(u,v)$ on $S$, what more work do I have to do? How do I find the equations for the asymptotic curves based on this angle?
If this whole method was for naught, how does one solve the differential equation. in this case, of: $$e (u')^2 + 2f u' v' + g (v')^2 = 2a v^4 (u')^2 - 2a u^3 v^3 \left( \frac{1}{u^2} + \frac{1}{v^2} \right)u' v' + 2a u^4 (v')^2 = 0?$$ (Again, assuming sufficient niceness.)
(See: https://math.stackexchange.com/questions/762195/differential-equation-for-the-asymptotic-directions-of-a-given-surface)
Thank you!
 [/INST] I don't have much time for elaborating now, but you definitely can do this yourself (feel free to add your own answer). 
Just observe that substitutions $x = r \cos \theta$ and $y = r \sin \theta$ to $z = a \left( x/y + y/x \right)$ turn this equation to $z = \tfrac{a}{\cos \theta \cdot \sin \theta}$ which clearly does not depend on $r$. This means that for each $\theta$ the ray $\theta = const$, $r>0$ lies on the surface. Now you can use Corollary 18.6. from Chapter 18 of A.Gray, E.Abbena and S.Salamon "Modern Differential Geometry of Curves and Surfaces with Mathematica" stating that "A straight line that is contained in a regular surface is necessarily an asymptotic curve".
Alternatively you may try to use the differential equation for asymptotic curves from here.
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