${e^y}\frac{{dy}}{{dx}} + x\frac{{dy}}{{dx}} + y = 0$
${e^y}\frac{{dy}}{{dx}} + x\frac{{dy}}{{dx}} + y = 0$
$\frac{{dy}}{{dx}}\left( {x + {e^y}} \right) = - y,{\left. {\frac{{dy}}{{dx}}} \right|_{\left( {0,1} \right)}} = - \frac{1}{e}$ again differentate w.r.t. $x$
${e^y}.\frac{{{d^2}y}}{{d{x^2}}} + \frac{{dy}}{{dx}}.{e^y}.\frac{{dy}}{{dx}} + x.\frac{{{d^2}y}}{{d{x^2}}} + \frac{{dy}}{{dx}} + \frac{{dy}}{{dx}} = 0$
$\left( {x + {e^y}} \right)\frac{{{d^2}y}}{{d{x^2}}} + {\left( {\frac{{dy}}{{dx}}} \right)^2}.{e^y} + 2\frac{{dy}}{{dx}} = 0$
$e\frac{{{d^2}y}}{{d{x^2}}} + \frac{1}{{{e^1}}}e + 2\left( { - \frac{1}{e}} \right) = 0$
$\therefore \frac{{{d^2}y}}{{d{x^2}}} + \frac{1}{{{e^2}}}$
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