Find the integrating factor of $(x^2y-2xy^2)\,dx+(x^3-3x^2y)\,dy=0$ 
Find the integrating factor of the differential equation: $$(x^2y-2xy^2)\,dx+(x^3-3x^2y)\,dy=0$$

What I tried:
This is a homogeneous equation.
Therefore,
$$I.F=\frac{1}{Mx+Ny}=\frac{1}{(x^2y-2xy^2)x+(x^3-3x^2y)y}=\frac{1}{x^3y-2x^2y^2+x^3y-3x^2y^2}$$
However, the given answer is:
$$I.F=\frac{1}{x^2y^2}$$
 A: I don't know if you apply the so-called "theorem" on convenient condition, but the equation that you found is false. Just check it with the correct integrating factor :
$$I.F.=x^2y^4$$
$$x^2y^4\left((x^2y-2xy^2)\,dx+(x^3-3x^2y)\,dy\right)=0$$
$$(x^4y^5-2x^3y^6)\,dx+(x^5y^4-3x^4y^5)\,dy=0$$
$$d(\frac15 x^5y^5-\frac12 x^4y^6)=0$$
$$\frac15 x^5y^5-\frac12 x^4y^6=C$$
A: Hint: Making the Substitution $$y(x)=xv(x)$$ then we get
$$\frac{dv(x)}{dx}=\frac{-5v(x)^2+2v(x)}{x(3v(x)-1)}$$ and this is
$$\frac{\frac{dv(x)}{dx}(3v(x)-1)}{-5v(x)^2+2v(x)}=\frac{1}{x}$$
and we can integrate
$$\int \frac{\frac{dv(x)}{dx}(3v(x)-1)}{-5v(x)^2+2v(x)}dx=\int \frac{1}{x}dx$$
Can you finish?
A: However I'm not familiar with your method, but perhaps this This method for homogeneous equation may help. We have
$$f(x,y)=-\dfrac{x^2y-2xy^2}{x^3-3x^2y}=y'$$
is homogeneous in sense of
$$\color{red}{f(tx,ty)=t^\alpha f(x,y)}$$
then 
$$y'=-\dfrac{\frac{y}{x}-2(\frac{y}{x})^2}{1-3\frac{y}{x}}$$
with $u=\dfrac{y}{x}$
$$u'x+u=-\dfrac{u-2u^2}{1-3u}$$
which is separable.
