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

Question

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

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

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

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

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

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