Given $e^{y}+x y=e$ On differentiating w.r.t. $x$, we get
$e^{y} \frac{d y}{d x}+y+x \frac{d y}{d x}=0$ ...(i)
At $x=0$ we get $e^{y}+0 . y=e \Rightarrow e^{y}=e \Rightarrow y=1$
$\therefore$ By putting $y=1$ in equation (i)
we get
$e \frac{d y}{d x}+1+0=0$
$\Rightarrow \frac{d y}{d x}=-\frac{1}{e}$
Again differentiating Eq. (i), we get
$e^{y} \frac{d^{2} y}{d x^{2}}+e^{y}\left(\frac{d y}{d x}\right)^{2}+\frac{d y}{d x}+x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=0$
$\Rightarrow \frac{d^{2} y}{d x^{2}}\left(e^{y}+x\right)+e^{y}\left(\frac{d y}{d x}\right)^{2}+\frac{2 d y}{d x}=0$
Now, At $x=0, y=1$
$\frac{d^{2} y}{d x^{2}}(e+0)+e\left(-\frac{1}{e}\right)^{2}+2\left(-\frac{1}{e}\right)=0$
$\Rightarrow e \frac{d^{2} y}{d x^{2}}-\frac{1}{e}=0$
$\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{1}{e^{2}}=e^{-2}$