How to solve a differential equation $(x^2y+y^5)dx+(x^3-xy^4)dy=0$? 
Solve the following differential equation:
  $$(x^2y+y^5)dx+(x^3-xy^4)dy=0$$

I noticed that it is not exact, since:
$$
(x^2y+y^5)'_y=x^2+5y^4\ne3x^2-y^4=(x^3-xy^4)'_x
$$
Then I tried to express $y'$:
$$
y'=\frac{x^2y+y^5}{xy^4-x^3}
$$
And now I don't understand what type of differential equation it is and how to solve it.
Could someone help me?

Probably, I can help a bit.
It seems to me, you can try substitution
$$
f = - \frac{1}{xy},\quad g = - \frac{y^3}{x^3}.
$$
Since
$$
df = \frac{dx}{x^2 y} + \frac{dy}{x y^2},\quad
dg = \frac{3 y^3 dx}{x^4} - \frac{3 y^2 dy}{x^3},
$$
you can write
$$
df + \frac{1}{3} dg = \frac{\left(x^2y + y^5\right)dx+ \left(x^3-x y^4\right)dy}{x^4 y^2}.
$$
Now solution should be straightforward.
