Multiplying both sides by [tex]y^2[/tex] gives
[tex]xy^2\dfrac{\mathrm dy}{\mathrm dx}+y^3=1[/tex]
so that substituting [tex]v=y^3[/tex] and hence [tex]\frac{\mathrm dv}{\mathrm dv}=3y^2\frac{\mathrm dy}{\mathrm dx}[/tex] gives the linear ODE,
[tex]\dfrac x3\dfrac{\mathrm dv}{\mathrm dx}+v=1[/tex]
Now multiply both sides by [tex]3x^2[/tex] to get
[tex]x^3\dfrac{\mathrm dv}{\mathrm dx}+3x^2v=3x^2[/tex]
so that the left side condenses into the derivative of a product.
[tex]\dfrac{\mathrm d}{\mathrm dx}[x^3v]=3x^2[/tex]
Integrate both sides, then solve for [tex]v[/tex], then for [tex]y[/tex]:
[tex]x^3v=\displaystyle\int3x^2\,\mathrm dx[/tex]
[tex]x^3v=x^3+C[/tex]
[tex]v=1+\dfrac C{x^3}[/tex]
[tex]y^3=1+\dfrac C{x^3}[/tex]
[tex]\boxed{y=\sqrt[3]{1+\dfrac C{x^3}}}[/tex]
