If e ^ y + xy = e, the ordered pair ((dy)/(dx), d ^ 2 y d x ^ 2 )...

${e}^{y} + xy = e$

differentiate w. r. t. $x$

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

⇒ \dfrac{dy}{dx}\left(x + {e}^{y}\right) = - y, {\left\dfrac{dy}{dx}\right|}_{( 0,1 )} = - \dfrac{1}{e}

Again differentiate w. r. t. $x$

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

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

Now, at \left(0, 1\right)$</p><p>$e\dfrac{{d}^{2}y}{d{x}^{2}} + \dfrac{1}{{e}^{2}}e + 2\left(- \dfrac{1}{e}\right) = 0$</p><p>$∴ \dfrac{{d}^{2}y}{d{x}^{2}} = \dfrac{1}{{e}^{2}}$</p><p>Hence, ordered pair \left(\dfrac{dy}{dx}, \dfrac{{d}^{2}y}{d{x}^{2}}\right) = \left(- \dfrac{1}{e}, \dfrac{1}{{e}^{2}}\right)$