Considering the surface $f(x,y)=x^2y$. We know that a parametrization it can be $X(u,v)=(u,v,u^2v)$. QUESTION: Considering the surface $f(x,y)=x^2y$. We know that a parametrization it can be $X(u,v)=(u,v,u^2v)$. Find the asymptotic lines in $S$.
MY ATTEMPT: Let $\alpha:I\subset \mathbb{R}\rightarrow S$ be a curve in this surface, such that $\alpha(t)=X(u(t), v(t))$ and $\alpha '(t)=u'X_u+v'X_v$. Thus, $\alpha $ is asymptotic if, and only if,
$$e(u')^2+2fu'v'+ g(v')^2=0. \qquad (*)$$
Where $e=\frac{2v}{\sqrt{1+u^4+4u^2v 2}}$, $f=\frac{2u}{\sqrt{1+u^4+ 4u^2v^2}}$ and $g=0$. Replacing this in $(*)$ we can find that $e(u')^2+2fu'v'+ g(v')^2=0\iff u'=0 \; \text{or} \; u'v+2uv'=0$. In the first case $u=\text{constant}$. However I'm struggling to resolve the ODE $u'v+2uv'=0$. Would you help me with this?

$$u'v+2uv'=0$$
$$\dfrac {u'}{u}=-2\dfrac{v'}{v}$$
Then write;  $$(\ln u)'=\dfrac {u'}{u} ;\; (\ln v)'=\dfrac {v'}{v}$$
$$(\ln u)'=-2(\ln v)'$$
Integrate.
