Dividing both sidesby \(y^{-2}\)

\(y^2y'+ y^3/x=3\)

\(u=y^3\)

Differentiating both sides with respect to x

\(du/dx= 3y^2 dy/dx\)

\(1/3 (du)/dx= y^2 dy/dx\)

Substituting this in the equation \(y^2y'+ y^3/x=3\)

Substituting \(1/3 (du)/dx= y^2 dy/dx\) and \(u=y^3\) in the equation \(y^2y'+ y^3/x=3\)

\(1/3 (du)/dx+ u/x=3\)

\(1/9 (du)/dx+u/(3x)=1\)