Question Let the funcion $z = z(x,y)$ be given by the equation $z = xy + f(x^2 -y)$, where $f$ is an arbitrary $C^1$ function. By forming the first partial derivatives of $z = z(x,y)$: $p = z_x$ and $q = z_y$, eliminate the unknown function $f$ by obtaining a PDE.
Attempt
$$ z_x = \frac{\delta z}{\delta x} = y + 2xf'(x^2 - y) = p $$ $$ z_y = \frac{\delta z}{\delta y} = x - f'(x^2 - y) = q $$
Moving things around
$$ \frac{y - p}{2x} = f'(x^2 - y) = x + q$$
giving us
$$ y - z_x = 2x^2 + 2x \cdot z_y$$
That's my answer. Is that right?
Also, where can I read more on this method and can you give me a little bit of background on the role of this in the grander scheme of things?
Warning with division for zero! However, $$ z_x+2xz_y=(y+2xf'(x^2-y))+2x(x-f'(x^2-y))=y+2x^2, $$ i.e., $$ z_x+2xz_y-y-2x^2=0. $$