Problem is to solve the differential equation $$y\frac{\partial f}{\partial x} -x \frac{\partial f}{\partial y}=x^3y+xy^3, x> 0, y> 0$$
using variables $u = x^2+y^2, v = x^2-y^2$
I have started at a solution with the chain rule, I have $$\frac{\partial f}{\partial x} = 2x\frac{\partial f}{\partial u}+ 2x\frac{\partial f}{\partial v} $$ $$\frac{\partial f}{\partial y} = 2y\frac{\partial f}{\partial u}-2y\frac{\partial f}{\partial v}$$ Plug in that in the differential equation I get $$y(2x\frac{\partial f}{\partial u}+2x\frac{\partial f}{\partial v})-x(2y\frac{\partial f}{\partial u}-2y\frac{\partial f}{\partial v})=x^3y+xy^3 \iff 4xy \frac{\partial f}{\partial v} = x^3y+xy^3 $$ But I am stuck here.
With cancellation you get
$$\frac{\partial f}{\partial v}=\frac{1}{4}(x^2+y^2)=\frac{1}{4}u$$ The right hand side is constant in $v$, and so it can be directly integrated, yielding $$f(u,v)=\frac{1}{4}uv+g(u)$$