Using integrating factors to solve difficult differential equations Consider the differential equatioN:
$$(2x^2 y -2y^4)dx+ ( 2x^3 + 3xy^3) dy = 0$$
This equation is of the form:
$$ Q dx +  P dy=0$$
Now,  it's easy to see that this differential is not exact by using the commutativity of second order partial condition. Multiply it by an integrating factor $\eta(x,y)$ such that it does:
$$ \eta_x  P + P_x \eta  = \eta_y Q + Q_y \eta$$
Or,
$$ \eta_x P - \eta_y Q + ( P_x - Q_y) \eta = 0$$
This becomes:(*)
$$ \eta_x ( 2x^3 + 3xy^3) - \eta_y ( 2x^2 y - 2y^4) + ( 6x^2 +3 y^3 - 2x^2 -8y^3) \eta =0$$
Or,
$$ \eta_x ( 2x^3 + 3xy^3) - \eta_y ( 2x^2 y - 2y^4) + ( 4x^2  - 11y^3) \eta =0$$
Now... I'm not quite sure what to do.. I don't think I can write $ \eta$ as function solely of $x$ or $y$.. did I miss something or... ?

As correctly observed  by @Aleksas Domarkas. There is a mistake after the place where I put *. The current working is as follows:
$$ \eta_x ( 2x^3 + 3xy^3) - \eta_y ( 2x^2 y - 2y^4) + ( 6x^2 +3 y^3 - 2x^2 +8y^3) \eta =0$$
Or,
$$ \eta_x ( 2x^3 + 3xy^3) - \eta_y ( 2x^2 y - 2y^4) + ( 4x^2  + 11y^3) \eta =0$$

$(2x^2y-2y^4)~dx+(2x^3+3xy^3)~dy=0$
$(2x^3+3xy^3)~dy=(2y^4-2x^2y)~dx$
$\dfrac{dy}{dx}=\dfrac{2y^4-2x^2y}{2x^3+3xy^3}$
Let $r=x^2$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{dr}\dfrac{dr}{dx}=2x\dfrac{dy}{dr}$
$\therefore2x\dfrac{dy}{dr}=\dfrac{2y^4-2x^2y}{2x^3+3xy^3}$
$\dfrac{dy}{dr}=\dfrac{y^4-x^2y}{x^4+3x^2y^3}$
$\dfrac{dy}{dr}=\dfrac{y^4-ry}{r^2+3ry^3}$
Let $s=y^3$ ,
Then $\dfrac{ds}{dr}=3y^2\dfrac{dy}{dr}$
$\therefore\dfrac{1}{3y^2}\dfrac{ds}{dr}=\dfrac{y^4-ry}{r^2+3ry^3}$
$\dfrac{ds}{dr}=\dfrac{3y^6-3ry^3}{r^2+3ry^3}$
$\dfrac{ds}{dr}=\dfrac{3s^2-3rs}{r^2+3rs}$
