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 $a$?
 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.
A: This is separable. The hint is in the terms outside the square root.
But first, lets talk a bit about where this is defined.
Clearly we must have $x \ge x^2y^2$. Which also means we must have $x\ge 0$. Furthermore, you can check that $x=0$ and $y=0$ are both solutions. So we will be looking for solutions in the $x>0$ region. 
Lets write the differential as
$$
y^2\sqrt{x(1-xy^2)}\ dx = ydx + 2xdy.
$$
Hopefully, you can see the the left hand side suggests that we can make the substitution $z = xy^2$ (check it), with $z\in (0,1)$. 
What comes out the other side is
$$
\frac{dx}{x} = \frac{dz}{\sqrt{z^3(1-z)}}.
$$
We can solve this to get
$$
\ln x  = -2\sqrt{\frac{1}{z} -1} +c.
$$
for some constant $c$.
Working our way back up we should find
$$
y = \pm \frac{2}{\sqrt{x}\sqrt{(c - \ln x)^2 +4)}}.
$$
