I want to solve for $f = f(\psi)$. $\psi = \psi(x,y)$ is a known function in $[0,2\pi] \times [0,2\pi]$.
I have the following equation:
$\partial_\psi f = - \frac{\nabla^2 \psi}{|\nabla \psi|^2}$
Where $\nabla$ is the gradient operator and $|\nabla \psi|$ is the Euclidean norm of $\nabla \psi$. $|\nabla \psi|$ is a smooth and positive function. $\nabla^2 \psi$ is the "Surface-Laplacian Operator" or simply the Laplacian operator for 2D functions.
So my idea is that I can integrate both sides of the equation with respecto to $\psi$ to obtain the following:
$f(\psi (x,y)) = - \int \frac{\nabla^2 \psi}{|\nabla \psi|^2} d\psi + C$, where $C$ is an arbitrary constant. Now $d\psi$ can be expanded as follows:
$d\psi = \partial_x \psi dx + \partial_y \psi dy$. Substituting in the integral equation we obtain the following:
$f(\psi) = - \int \frac{\nabla^2 \psi}{|\nabla \psi|^2} (\partial_x \psi dx + \partial_y \psi dy) + C$
Am I on the right track after integrating both sides with respect to $\psi$ and expanding the "function differential" as I did?