It's unclear whether the ODE is
[tex]xy'+y=4xy^2[/tex]
or
[tex]xy+y'=4xy^2[/tex]
If the first case, then divide through both sides by [tex]y^2[/tex] to get
[tex]xy^{-2}y'+y^{-1}=4x[/tex]
then substitute [tex]z=y^{-1}[/tex], so that [tex]z'=-y^{-2}y'[/tex]. Then
[tex]-xz'+z=4x\implies-\dfrac1xz'+\dfrac1{x^2}z=\left(-\dfrac1xz\right)'=\dfrac4x[/tex]
[tex]\implies-\dfrac1xz=\displaystyle\int\frac4x\,\mathrm dx[/tex]
[tex]\implies z=-4x\ln|x|+Cx[/tex]
[tex]\implies\boxed{y=\dfrac1{Cx-4x\ln|x|}}[/tex]
If the second, then dividing through by [tex]y^2[/tex] gives
[tex]xy^{-1}+y^{-2}y'=4x[/tex]
The same substitution as before gives
[tex]xz-z'=4x\implies e^{-x^2/2}z'-xe^{-x^2/2}z=\left(e^{-x^2/2}z\right)'=-4xe^{-x^2/2}[/tex]
[tex]\implies e^{-x^2/2}z=\displaystyle-4\int xe^{-x^2/2}\,\mathrm dx[/tex]
[tex]\implies z=4+Ce^{x^2/2}[/tex]
[tex]\implies\boxed{y=\dfrac1{4+Ce^{x^2/2}}}[/tex]
