Integrating factor for DE of the form DE $f(xy) ydx+g(xy) xdy=0$

Question

I know that IF is of the form $$ \frac{1}{Mx-Ny} $$

But I couldn't find a formal proof anywhere. So far I could do the following:

$f(xy)x=P$ and $g(xy)y=Q$

$(\mu P)_y = (\mu Q)_x$

$f\mu + y f_y\mu+ xyf\mu_y = g\mu + xf_x\mu+xyf\mu_x$

Answer 1

$$u ( f(u)-g(u) ) dx + xg(u) du = 0$$ Divide by the integrating factor :$$\mu=xu ( f(u)-g(u) )=x^2y ( f(xy)-g(xy) )$$ $$\mu =x(xM-yN)$$

Then the DE becomes exact. $$\dfrac {dx}{x}+ \dfrac {g(u)} {u ( f(u)-g(u) )} du = 0$$