How to solve such fraction differential equation? Here's my first-order differential equation:
$$
(x^3 - 2xy^2)dx + 3yx^2dy = xdy - ydx
$$
I've tried to make it fraction, but it isn't separable differential equation, also it isn't differential equation in total differentials, so after it I lose any clue for answer.

$$(x^3 - 2xy^2)dx + \color {red}{ 3yx^2dy} = xdy - ydx$$
$$(x^3 - 2xy^2)dx +\color {red}{ 2yx^2dy+yx^2dy} = xdy - ydx$$
Rearrange terms:
$$(x^3dx+x^2ydy) - 2xy(ydx - xdy )= xdy - ydx$$
$$x^2(xdx+ydy) = (xdy - ydx)(1-2xy)$$
$$xdx+yd y = (1-2xy)d  \left (\dfrac {y}{x}\right)$$
$$d(x^2+y^2)  = 2(1-2xy)d  \left (\dfrac {y}{x}\right)$$
Divide by $x^2+y^2$ and substitute $u=x^2+y^2$ and $v=\dfrac yx$:
$$\dfrac 12 \dfrac {du}{u} =\left (\dfrac 1u-\dfrac {2v}{v^2+1}\right)dv$$
$$ \dfrac {du}{dv} = 2-\dfrac {4uv}{v^2+1}$$
$$ u' +\dfrac {4v}{v^2+1}u=2$$
This is a first order linear DE. Try to integrate by integrating factor method:
$$\mu (v)=\exp \int \dfrac {4v }{v^2+1}dv$$
$$\mu (v)=\exp \int \dfrac {2}{v^2+1} dv^2=(v^2+1)^2$$
The DE becomes:
$$ (u(v^2+1)^2)'=2(v^2+1)^2$$
$$ u(v^2+1)^2=2\left(\dfrac {v^5}{5}+2\dfrac {v^3}{3}+v\right)+C$$
Unsubstitute $u$ and $v$:
$$u=x^2+y^2 \text { and }v=\dfrac yx$$
