1. The problem statement
Find a solution of $$\frac{1}{x^2}\frac{\partial u(x,y)}{\partial x}+\frac{1}{y^3}\frac{\partial u(x,y)}{\partial y}=0$$ Which satisfies the condition $\frac{\partial u(x,y)}{\partial x}\big |_{y=0}=x^3$ for all $x$.
2.The attempt at a solution
I get the following general solution for the PDE:
$$u(x,y)=f(\frac{x^3}{3}-\frac{y^4}{4}):=f(z)$$
For some function f which satisfies the BC. I'm not too sure how to deal with this type of boundary condition. But here is an attempt:
$\partial_x u=\frac{\text{d}f}{\text{d}z}\frac{\partial z}{\partial x}=\frac{\text{d}f}{\text{d}z} x^2$. Thus $\frac{\text{d}f}{\text{d}z} x^2=x^3\implies \frac{\text{d}f}{\text{d}z} =x\implies f(z)=zx$
At $y=0,~ z=\frac{x^3}{3}\implies x=(3z)^{\frac{1}{3}}$. So $f(z)=(3z)^{\frac{1}{3}}z$.
Changing variables back gives:
$u(x,y)=(3(\frac{x^3}{3}-\frac{y^4}{4}))^{\frac{1}{3}}(\frac{x^3}{3}-\frac{y^4}{4})=3^{\frac{1}{3}}(\frac{x^3}{3}-\frac{y^4}{4})^{\frac{4}{3}}$
This is a solution to the PDE but doesn't satisfy the given boundary condition, it is off by a factor 3/4 and i'm not sure why.
If $u (x,y) = f \left(\frac{x^3}{3} - \frac{y^4}{4}\right)$, then the boundary condition gives us $f ' \left(\frac{x^3}{3}\right) = x$. Hence, we have
$$f ' (z) = \sqrt[3]{3z}$$
Integrating, we obtain
$$f (z) = \frac{3\sqrt[3]{3}}{4} \, z^{\frac{4}{3}}$$
and, thus,
$$u (x,y) = \frac{3\sqrt[3]{3}}{4} \left(\frac{x^3}{3} - \frac{y^4}{4}\right)^{\frac{4}{3}}$$