Consider $u:\Omega\subset \mathbb R^2\to \mathbb R$ be a $C^2$ function. Knowing that $\nabla u\neq 0$ everywhere on $\Omega$, $\Delta u\equiv 1$ and $|\nabla u(x,y)|=f(u(x,y))$ for some $f$ $C^1$, prove that, locally, the level curves of $u$ are lines or circles.
My attempt was to prove that these curves have locally constant curvature. I calculated the curvature, following the Frenet formulas, as the module of the derivative of the normal vector. So, given a local arc length parametrization of a level curve (it exists by the implicit function theorem) we can say that $$k(t)=\Big|\frac{d}{dt}\frac{\nabla u(x(t),y(t))}{f(u(x(t),y(t)))}\Big|$$ since the gradient is normal to the level curve.
Developing the formulas, one actually makes use of the hypotesis $\Delta u=1$ in order to simplify the expression, but I didn't obtain that $k$ is constant.
Is my attempt successful? Are there any better ways?
Thank you in advance.
Let $N=\nabla u/|\nabla u|$ be the unit normal to a level set of $u$ and $T = R_{\pi/2}N$ the unit tangent, and use the formula
$$ k = -\frac{\nabla^2 u (T, T)}{|\nabla u|} $$ for the curvature (if you have trouble deriving this you can find it in e.g. Prop 3.1 here).
Since $$\Delta u = \text{tr} \nabla^2 u = \nabla^2 u(T,T) + \nabla^2 u(N,N) = 1,$$ we can write this as
$$k = \frac{\nabla^2 u (N,N) - 1}{|\nabla u|} = \frac{\nabla^2 u (\nabla u, \nabla u)-|\nabla u|^2}{|\nabla u|^3}.$$
Now note that since $\nabla^2 u \cdot \nabla u = \frac12 \nabla |\nabla u|^2 = \frac12 \nabla (f(u)^2)=f(u) f'(u) \nabla u$, we have $$\nabla^2u (\nabla u, \nabla u) = f(u)f'(u)|\nabla u|^2 = f(u)^3 f'(u),$$
so the curvature is $$k=f'(u) - \frac 1 {f(u)},$$ which depends only on $u$ and is thus constant on the level sets.