If $e^y+xy=e,$ then the value of $\frac{d^{2}y}{d{x}^2}$ for $x=0,$ is

We have $e^y+xy=e$. Differentiating w.r.t.x, we get $e^y\frac{dy}{dx}+y+x\frac{dy}{dx}=0.....\left(i\right)$ 

Differentiating w.r.t.x, we get $e^y\frac{d^{2}y}{d{x}^2}+e^y{\left(\frac{dy}{dx}\right)}^2+2\frac{dy}{dx}+x\frac{d^{2}y}{d{x}^2}=0.....\left(ii\right)$

$\text{Putting }x=0 \text{in }{e}^y+xy=e\text{, we get }y=1$

$\text{Putting }x=0, y=1 \text{in (i), we get }e\frac{dy}{dx}+1=0\Rightarrow \frac{dy}{dx}=-\frac{1}{e}$

$\text{Putting }x=0, y=1, \frac{dy}{dx}=-\frac{1}{e} \text{in (ii), we get }e\frac{d^{2}y}{d{x}^2}+e\cdot \frac{1}{e^2}-\frac{2}{e}+0$ $\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{1}{e^2}$