[tex]xy''=y'+(y')^3[/tex]
If [tex]u=y'[/tex], then [tex]u'=y''[/tex] and we have
[tex]xu'=u+u^3[/tex]
Divide both sides by [tex]u^3[/tex], so that we get
[tex]xu^{-3}u'=u^{-2}+1[/tex]
This is the Bernoulli equation we wanted to find. So we can substitute [tex]w=u^{-2}[/tex]. This gives [tex]w'=-2u^{-3}u'[/tex], so the ODE is equivalent to
[tex]-\dfrac12xw'=w+1[/tex]
which is linear in [tex]w[/tex]. Multiply both sides by [tex]x[/tex] and rearrange terms a bit to get
[tex]\dfrac12xw'+w=-1\implies x^2w'+2xw=-2x[/tex]
and now the LHS contains a derivative of a product, namely
[tex](x^2w)'=-2x\implies x^2w=-x^2+C\implies w=Cx^{-2}-1[/tex]
Now you can solve for [tex]u[/tex], then integrate that result to find [tex]y[/tex].
