First order ODE $(y^2\sqrt{x-x^2y^2}-y)dx - 2xdy=0$ $(y^2\sqrt{x-x^2y^2}-y)dx - 2xdy=0$
Change it into this $$y'=\frac{y^2\sqrt{x-x^2y^2}-y}{2x}$$
Square root disables a lot of methods. 
It isn't a total differential, I've tried. Quasi-homogeneous  $y=z^a$ is only thing I haven't ruled out but I don't know how to determine $a$ because of the square root. If it is quasi-homogeneous how do I find $a$?

All first order equation of the first $M(x,y)dx+N(x,y)dy=0$ can be integrated, provided that you can find the right integrating factor. For this equation $(y^2\sqrt{x-x^2y^2}-y)dx - 2xdy=0$, the integrating factor is 
$$ \frac{1}{xy^2\sqrt{x-x^2y^2}}.$$
Multiplying both sides with this integrating factor, we get
$$ \left(\frac{1}{x}-\frac{1}{xy\sqrt{x-x^2y^2}}\right)dx-\frac{2}{y^2\sqrt{x-x^2y^2}}dy=0.$$
This is now a total differential and integrating on both sides leads to
$$ \ln x-\frac{2}{y}\sqrt{\frac{1-xy^2}{x}}=C.$$
The solution then can be obtained by solving $y$ from the algebraic equation.
