Tough Differential equation. Please help!

dy/dx-y/x=y^3

I just can't figure it out. It doesn't make sense to me. I feel like it's a substitution but I don't know where to go.

thanks for your help.

dy/dx = y³ + y/x
dy/dx = (xy³+y)/x
dy/dx = y(xy²+1)/x

Let u = xy²

Differentiate both sides with respect to x

du/dx = y² + 2xy dy/dx
du/dx = y² + 2xy * y(xy²+1)/x
du/dx = y² + 2xy²(xy²+1)/x ------> xy² = u -----> y² = u/x
du/dx = u/x + 2u(u+1)/x
du/dx = (u+2u(u+1))/x
du/dx = (2u²+3u)/x
1/(2u²+3u) du = 1/x dx
3/(u(2u+3)) du = 3/x dx
(1/u − 2/(2u+3)) du = 3/x dx

Integrate both sides:
ln|u| − ln|2u+3| + ln|C| = 3 ln|x|
ln|Cu/(2u+3)| = ln|x³|
Cu/(2u+3) = x³

Cxy²/(2xy²+3) = x³
Cy²/(2xy²+3) = x²
Cy² = 2x³y² + 3x²
y² (C − 2x³) = 3x²
y² = 3x² / (C − 2x³)

This may be against the Q&A rule but holy c**p that's hard (well im only in grade 7 so it's not saying a lot)