Continuing on my path to learn differential geometry, I need help with the following exercise from the theory of surfaces:
Consider the surface $S : \vec x = \vec x(u, v) \in C^3$ where
$$\vec n_u = a(u, v) \vec x_u, \quad \vec n_v = a(u, v) \vec x_v \tag{*}$$
and $a(u, v) \in C^2$ is a nonvanishing function. Show that $S$ is a sphere.
We use $\vec x(u, v)$ to denote the parametric representation of the surface $S$ with parameters $u$ and $v$ and it's partial derivatives by $\vec x_u = \frac{\partial \vec x}{\partial u}$ and $\vec x_v = \frac{\partial \vec x}{\partial v}$. Also, we denote by $\vec n$ the unit normal vector to the surface at a point $P$. Finally, a surface $S$ is said to be of class $C^n$ if it has partial derivatives of all orders to $n$.
My first thought was to prove that the position vector at a random point of the surface has constant norm but, considering the informations we're given, I don't think this is possible. I think I'm missing some key points about the properties of a sphere. Obviously the equations $(*)$ are important, but I don't see why.
Step I: $a$ is constant.
Differentiate the first with respect to $v$. $$N_{uv}=a_vx_u+ax_{uv}$$ and the second with respect to $u$,
$$N_{vu}=a_ux_v+ax_{vu}.$$
Since mixed partials are equal we have
$$a_vx_u=a_ux_v$$ but since $x_u$ and $x_v$ are independent, $a_u=a_v=0$.
Step II $y=x-\frac{1}{a}n$ is constant.
Step III $|x-y|=\frac{1}{a}$.
Note that if $a=0$ you get a plane.