Let y = y ( x ) be the solution of the differential equation...

Let $y + 1 = Y$

$∴ \dfrac{dY}{dx} = {Y}^{2}e\dfrac{{x}^{2}}{2} - xY$

Put $- \dfrac{1}{Y} = k$

$⇒ \dfrac{dk}{dx} + k\left(- x\right) = e\dfrac{{x}^{2}}{2}$

$I . F . = {e}^{- \dfrac{{x}^{2}}{2}}$

$∴ k = \left(x + c\right){e}^{{x}^{2} / 2}$

Put $k = - \dfrac{1}{y + 1}$

$∴ y + 1 = - \dfrac{1}{( x + c ) {e}^{{x}^{2} / 2}} . . . \left(i\right)$

when $x = 2, y = 0$, then $c = - 2 - \dfrac{1}{{e}^{2}}$

Differentiate equation $( i ) \text{ and }$ put $x = 1$ we get,

${\left(\dfrac{dy}{dx}\right)}_{x = 1} = - \dfrac{{e}^{3 / 2}}{{\left(1 + {e}^{2}\right)}^{2}}$