If $e^y+xy=e,$ the ordered pair $\left(\frac{dy}{dx},\frac{d^{2}y}{d{x}^2}\right)$ at $x=0$ is equal to

$e^y+xy=e$

differentiate w. r. t. $x$

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

$\Rightarrow \frac{dy}{dx}\left(x+e^y\right)=-y, {\left.\frac{dy}{dx}\right|}_{\left(\mathrm{0,1}\right)}=-\frac{1}{e}$

Again differentiate w. r. t. $x$

$e^y.\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}.e^y.\frac{dy}{dx}+x.\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+\frac{dy}{dx}=0$

$\Rightarrow \left(x+e^y\right)\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2.e^y+2\frac{dy}{dx}=0$

Now, at $\left(0,1\right)$

$e\frac{d^{2}y}{d{x}^2}+\frac{1}{e^2}e+2\left(-\frac{1}{e}\right)=0$

$\therefore \frac{d^{2}y}{d{x}^2}=\frac{1}{e^2}$

Hence, ordered pair $\left(\frac{dy}{dx},\frac{d^{2}y}{d{x}^2}\right)=\left(-\frac{1}{e},\frac{1}{e^2}\right)$