If $e^y+xy=e$ , then $y_2\left(0\right)=$

$e^y+xy=e$

$e^y\frac{dy}{dx}+x\frac{dy}{dx}+y=0$

$\frac{dy}{dx}\left(e^y+x\right)=-y$

$\Rightarrow \frac{dy}{dx}=\frac{-y}{e^y+x}$

$\frac{d^{2}y}{d{x}^2}=\frac{-\left[\left(e^y+x\right)\frac{dy}{dx}-y\left(e^y\frac{dy}{dx}+1\right)\right]}{{\left(e^y+x\right)}^2}$

$y_2\left(0\right)=\frac{-e\left(-\frac{1}{e}\right)+1\left(e\left(\frac{-1}{e}\right)+1\right)}{e^2}$                                $\left(\because \frac{dy}{dx/atx=e}=\frac{-1}{e}\right)$

$=\frac{0+1}{e^2}$

$=\frac{1}{e^2}$