I'm not sure. The solutions given are very vague. I think this is the solution to the differential equation.
Thanks.
$\displaystyle \frac{y'}{y^3}+\frac1{y^2}=x,$ 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.