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

Question

$\left(\frac{1}{e}, \frac{1}{e^2}\right)$

$\left(\frac{-1}{e}, \frac{-1}{e^2}\right)$

$\left(\frac{1}{e}, \frac{-1}{e^2}\right)$

$\left(\frac{-1}{e}, \frac{1}{e^2}\right)$

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