Integrating Factor by Inspection $(x^3+xy^2+y)dx + (y^3+xy^2+x)dy=0$ $$(x^3+xy^2+y) \hspace{.1cm} dx + (y^3+xy^2+x)\hspace{.1cm} dy=0$$ So I tried to solve this problem but can't figure out my integrating factor all I can see here is if I distribute first I can get a $y \hspace{.1cm} dx + x dy$ so that would be $d(xy)$ however I can't integrate the $xy^2dx$ and $xy^2dy$. Help please and can you also give me techniques on how to know the integrating Factors? My professors just told me that it's a trial and error process, is that true? 

Hint:
$(x^3+xy^2+y)~dx+(y^3+xy^2+x)~dy=0$
$\dfrac{dx}{dy}=-\dfrac{y^3+xy^2+x}{x^3+xy^2+y}$
With reference to http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=180:
Let $u=\dfrac{x}{y}$ ,
Then $x=yu$
$\dfrac{dx}{dy}=y\dfrac{du}{dy}+u$
$\therefore y\dfrac{du}{dy}+u=-\dfrac{y^3+y^3u+yu}{y^3u^3+y^3u+y}$
$y\dfrac{du}{dy}=-\dfrac{y^3(u+1)+yu}{y^3(u^3+u)+y}-u$
$y\dfrac{du}{dy}=-\dfrac{y^3(u^4+u^2+u+1)+2yu}{y^3(u^3+u)+y}$
$\dfrac{du}{dy}=-\dfrac{y^2(u^4+u^2+u+1)+2u}{y^3(u^3+u)+y}$
Let $v=\dfrac{1}{y^2}$ ,
Then $\dfrac{du}{dy}=\dfrac{du}{dv}\dfrac{dv}{dy}=-\dfrac{2}{y^3}\dfrac{du}{dv}$
$\therefore-\dfrac{2}{y^3}\dfrac{du}{dv}=-\dfrac{y^2(u^4+u^2+u+1)+2u}{y^3(u^3+u)+y}$
$\left(\dfrac{2}{y^2}+u^3+u+1+\dfrac{1}{u}\right)\dfrac{dv}{du}=\dfrac{2}{uy^4}+\dfrac{2(u^2+1)}{y^2}$
$\left(2v+u^3+u+1+\dfrac{1}{u}\right)\dfrac{dv}{du}=\dfrac{2v^2}{u}+2(u^2+1)v$
This belongs to an Abel equation of the second kind.
