Stuck at solving differential equation by using integrating factor This is the text:
Solve the following differential eqution: $$(2x^3+3x^2y+y^2-y^3)~dx+(2y^3+3x^2y+x^2-x^3)~dy=0$$
This particular differential equation has to be solved by finding an integrating factor. I first tried making this equation into a Darboux differential eqaution, and didn't find any luck. These are the integrating factors that I tried: $u(x,y)=u(x)$ , $u(x,y)=x^my^n$ , $u(x,y)=u(y)$ .
I don't have any idea what else to try. You don't have to answer it fully, just give me a hint or something. Thank you in advance.

$(2x^3+3x^2y+y^2-y^3)~dx+(2y^3+3x^2y+x^2-x^3)~dy=0$
$\dfrac{dy}{dx}=-\dfrac{2x^3+3x^2y+y^2-y^3}{2y^3+3x^2y+x^2-x^3}$
Let $y=xu$ ,
Then $\dfrac{dy}{dx}=x\dfrac{du}{dx}+u$
$\therefore x\dfrac{du}{dx}+u=-\dfrac{2x^3+3x^3u+x^2u^2-x^3u^3}{2x^3u^3+3x^3u+x^2-x^3}$
$x\dfrac{du}{dx}=\dfrac{(u^3-3u-2)x-u^2}{(2u^3+3u-1)x+1}-u$
$x\dfrac{du}{dx}=\dfrac{(-2u^4+u^3-3u^2-2u-2)x-u^2-u}{(2u^3+3u-1)x+1}$
$((2u^4-u^3+3u^2+2u+2)x+u^2+u)\dfrac{dx}{du}=-(2u^3+3u-1)x^2-x$
Let $x=\dfrac{1}{v}$ ,
Then $\dfrac{dx}{du}=-\dfrac{1}{v^2}\dfrac{dv}{du}$
$\therefore-\left(\dfrac{2u^4-u^3+3u^2+2u+2}{v}+u^2+u\right)\dfrac{1}{v^2}\dfrac{dv}{du}=-\dfrac{2u^3+3u-1}{v^2}-\dfrac{1}{v}$
$((u^2+u)v+2u^4-u^3+3u^2+2u+2)\dfrac{dv}{du}=v^2+(2u^3+3u-1)v$
This belongs to an Abel equation of the second kind.
