How to change this equation into one with separable variations?

Question

Consider $$\frac{dy}{dx}=\frac{2x^3+3xy^2+x}{3x^2y+2y^3-y}.$$ First RHS is not homogeneous, the $x$ and $y$ terms of order 1 are very annoying. And I noticed that $2x^3+3xy^2+x=(x+y)^3-3x^2y-y^3+x^3+x$, but no further. I tried substitution with $z=y/x^n$, but with no luck. I hope to find a substitution to change it into an equation with separable variables. Thank you in advance.
ED1:The Given solution is $(y^2-x^2+2)^5=c(x^2+y^2)$. But I care about the substitution.

Answer 1

You can factorize numerator and denominator as $$ \frac{dy}{dx}=\frac{x(2x^2+3y^2+1)}{y(3x^2+2y^2-1)} $$ Change the variables to $u=x^2-1$ and $v=y^2+1$ to get $$ \frac{dv}{du}={}^{''}\left(\frac{y\,dy}{x\,dx}\right){}^{''} =\frac{2(x^2-1)+3(y^2+1)}{3(x^2-1)+2(y^2+1)} =\frac{2u+3v}{3u+2v} $$ which is a standard homogeneous (fraction) differential equation.

Answer 2

$$\frac{dy}{dx}=\frac{2x^3+3xy^2+x}{3x^2y+2y^3-y}.$$ $$\frac{dy^2}{dx^2}=\frac{2x^2+3y^2+1}{3x^2+2y^2-1}.$$ The substitution is obvious to get the classical DE: $$\frac{du}{dv}=\frac{2u+3v+1}{3u+2v-1}$$. Where $u=y^2$ and $v=x^2$.