Solve $x\ y\ dx=(y^3+x^2y+x^2)\ dy$ differential equation. Solve $x\ y\ dx=(y^3+x^2y+x^2)\ dy$ differential equation.
What I tried so far.
$M(x,y)=x\ y$ $N(x,y)=y^3+x^2y+x^2$
$\frac{\partial M}{\partial y}=x$ $\frac{\partial N}{\partial x}=2x+2xy$
They are not equal so we need integrating factor.
$$\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}  }{N}=\frac{2xy-x}{y^3+x^2y+x^2}$$
so $$\mu(x)=e^{-{\huge \int}{\frac{(2xy-x)dx}{y^3+x^2+x^2y}}}$$
from here I can't continue.
EDIT.
I am getting $$\mu=|(1+y)\ x^2+y^3|^\frac{1-2y}{2+2y}$$

$$xydx=(y^3+x^2y+x^2)dy$$
$$x(ydx-xdy)=y(y^2+x^2)dy$$
$$(ydx-xdy)=\dfrac yx(y^2+x^2)dy$$
Divide by $y^2$:
$$d \left (\dfrac  {x}y \right)=\dfrac y x \left(1+\dfrac {x^2}{y^2} \right)dy$$
This is separable:
$$\dfrac  {du}{1+u}=2dy$$
With $u=\left (\dfrac  xy\right)^2$.
