Problem: $\displaystyle x^4(x+y)=y^2(3x-y)$

Solution attempt:

$\displaystyle \frac{d}{dx}[x^4(x+y)]-\frac{d}{dx}[y^2(3x-y)]=0$

$\displaystyle [4x^3(x+y)+x^4(1+y')] - [2yy'(3x-y)+y^2(3y-y')]=0$

upon distribution of coefficients,

$\displaystyle 4x^4+4x^3y+x^4+x^4y'-6xyy'+2y^2y'-3y^2+y^2y'=0 $

If I didn't subtract $\displaystyle \frac{d}{dx}[y^2(3x-y)]$ as I did in step 1 I end up with,

$\displaystyle 4x^4+4x^3y+x^4+x^4y'=6xyy'+2y^2y'-3y^2+y^2y' $

This is where I am confronted with many mathematically valid yet counter productive steps as I try to solve for $\displaystyle y'$

combining like terms I have,

$\displaystyle 5x^4+4x^3y+x^4y'=6xyy'+3y^2y'-3y^2$

$\displaystyle 5x^4+4x^3y+x^4y'=y'(6xy+3y^2)-3y^2$

Answer in book: $\displaystyle y'=\frac{3y^2-5x^4-4x^3y}{x^4+3y^2-6xy}$

notes: what gets me is the fact that I have two $\displaystyle y'$ and $\displaystyle \frac{y'}{y'}=1$