I have been trying to solve the following problem, but I am getting stuck...
we have: $-y^2dx + x^2dy = 0$
given: $u(x,y) = \frac{1}{(x-y)^2}$
both have partial derivative: $\frac{-2xy}{(x-y)^3}$
so we have the exact equation: $\frac{-y^2}{(x-y)^2}dx + \frac{x^2}{(x-y)^2}dy = 0$
now, I try integrate the dx expression: $\int\frac{-y^2}{(x-y)^2}dx$
This gives me: $\frac{y^2}{(x-y)} + g(y)$
Differentiating with respect to y gives: $\frac{-y(y-2x)}{(y-x)^2} +g'(y)$
I have checked my work for the above integral and derivative and also used an online calculator to verify that they are correct. Every other problem of this type that I have tried has had something meaningful come of:
$\frac{-y(y-2x)}{(y-x)^2} +g'(y) = \frac{x^2}{(x-y)^2}$
However, I have not been able to get anywhere using this to try to find a value for $g(y)$
The point is that now you calculate $g$, so you solve for $g'(y)$, getting $\frac{x^2+y(y-2x)}{(x-y)^2}=1$ so $g(y)$ can be chosen to be $y$, so the implicit solution is $\frac{y^2}{x-y}+y=C$, assuming your prior steps were correct.