Multiply both sides by [tex]x^{-3}[/tex] to get a homogeneous equation.
[tex](2xy^2 - x^3) \, dy + (y^3 - 2yx^2) \, dx = 0 \\\\ \implies \left(2\dfrac{y^2}{x^2} - 1\right) \, dy + \left(\dfrac{y^3}{x^3} - 2\dfrac yx\right) \, dx = 0[/tex]
Substitute [tex]y=vx[/tex] and [tex]dy = x\,dv + v\,dx[/tex]. This makes the equation separable.
[tex](2v^2 - 1) (x\,dv + v\,dx) + (v^3 - 2v) \, dx = 0[/tex]
[tex]x (2v^2 - 1) \,dv + (2v^3 - v) \, dx + (v^3 - 2v) \, dx = 0[/tex]
[tex]x (2v^2 - 1) \,dv + (3v^3 - 3v) \, dx = 0[/tex]
Separate the variables.
[tex]x (2v^2 - 1) \, dv = (3v - 3v^3) \, dx[/tex]
[tex]\dfrac{2v^2-1}{3v-3v^3} \, dv = \dfrac{dx}x[/tex]
[tex]\dfrac13 \dfrac{1 - 2v^2}{v(v-1)(v+1)} \, dv = \dfrac{dx}x[/tex]
Expand the left side into partial fractions.
[tex]\dfrac{1-2v^2}{v(v-1)(v+1)} = \dfrac av + \dfrac b{v-1} + \dfrac c{v+1} \\\\ \dfrac{1-2v^2}{v(v-1)(v+1)} = \dfrac{a(v^2-1) + bv(v+1) + cv(v-1)}{v(v-1)(v+1)} \\\\ 1-2v^2 = (a+b+c) v^2 + (b-c) v - a[/tex]
Solve for the coefficients [tex]a,b,c[/tex].
[tex]\begin{cases} a + b + c = -2 \\ b-c = 0 \\ -a = 1 \end{cases} \implies a=-1, b=c=-\dfrac12[/tex]
Thus our equation becomes
[tex]\left(-\dfrac1{3v} - \dfrac1{6(v-1)} - \dfrac1{6(v+1)}\right) \, dv = \dfrac{dx}x[/tex]
Integrate both sides.
[tex]\displaystyle \int \left(-\dfrac1{3v} - \dfrac1{6(v-1)} - \dfrac1{6(v+1)}\right) \, dv = \int \dfrac{dx}x[/tex]
[tex]\displaystyle -\frac13 \ln|v| - \frac16 \ln|v-1| - \frac16 \ln|v+1| = \ln|x| + C[/tex]
Solve for [tex]v[/tex] (as much as you can, anyway).
[tex]\displaystyle -\frac16 \left(2\ln|v| + \ln|v-1| + \ln|v+1|\right) = \ln|x| + C[/tex]
[tex]\displaystyle \ln\left|\frac1{\sqrt[6]{v^2(v^2-1)}}\right| = \ln|x| + C[/tex]
[tex]\displaystyle \frac1{\sqrt[6]{v^4-v^2}} = Cx[/tex]
[tex]\sqrt[6]{v^4-v^2} = \dfrac Cx[/tex]
[tex]\left(\sqrt[6]{v^4-v^2}\right)^6 = \left(\dfrac Cx\right)^6[/tex]
[tex]v^4-v^2 = \dfrac C{x^6}[/tex]
Put the solution back in terms of [tex]y[/tex].
[tex]\left(\dfrac yx\right)^4-\left(\dfrac yx\right)^2 = \dfrac C{x^6}[/tex]
[tex]y^4 - x^2y^2 = \dfrac C{x^2}[/tex]
[tex]\boxed{x^2y^4 - x^4y^2 = C}[/tex]
which is about as simple as we can hope to get this.
