To Solve: $\displaystyle py^3+qx^2=0$
where $p = \dfrac{\partial z}{\partial x}$, $q = \dfrac{\partial z}{\partial y}$.
My attempt:
Let $\displaystyle z=X(x)Y(y)$. So, $\displaystyle X'Yy^3+XY'x^2=0$
Separating the variables, $\displaystyle \frac{X'}{Xx^2}=\frac{-Y'}{Yy^3}$
Now how do I integrate this ? $X$ is an unknown function and $x$ is a variable. I don't know how to proceed?
You've got
$$\dfrac{X'}{Xx^2} = -\dfrac{Y'}{Yy^3}$$
where the LHS cannot contain any terms of $y$ (as $X$ is purely a function of $x$), and the RHS, similarly, cannot contain any terms of $x$. But as the two are equal, this implies that they cannot contain any terms of $x$ or $y$ $-$ in other words, they are equal to a constant (say $k$). Thus:
$$\dfrac{X'}{Xx^2} = -\dfrac{Y'}{Yy^3} = k \Rightarrow\\ X' = kXx^2,\ -Y' = kYy^3 \Rightarrow\\ \boxed{X' - kx^2X = 0\\ Y' + ky^3Y = 0} $$
These are ODEs that can be easily solved (and $k$ is an arbitrary constant).