Plz help me

To do this you need to make the substitution $\displaystyle \begin{align*} z = y^{-1} \end{align*}$ . Then $\displaystyle \begin{align*} y = z^{-1} \implies \frac{dy}{dx} = -z^{-2}\,\frac{dz}{dx} \end{align*}$ . Then your DE becomes

$\displaystyle \begin{align*} \frac{dy}{dx} + x\,y &= x\,y^2 \\ -z^{-2}\,\frac{dz}{dx} + x\, z^{-1} &= x\,z^{-2} \\ \frac{dz}{dx} - x\,z &= -x \end{align*}$

which is now linear.