\(\displaystyle \frac{y'}{y^3}+\frac1{y^2}=x,\) if \(\displaystyle u=\frac1{y^2}\implies u'=-\frac{2y'}{y^2},\) then \(\displaystyle -\frac12u'+u=x\implies u'-2u=-2x,\) now \(\displaystyle e^{-2x}u'-2e^{-2x}u=-2xe^{-2x},\) and then \(\displaystyle \big(u(x)e^{-2x}\big)'=-2xe^{-2x}.\)

you can solve the rest now.