Official solution
y = 1 \[ \Rightarrow \] x = 0
\[{e^y}{{dy} \over {dx}} + x{{dy} \over {dx}} + y = 0\]
\[ \Rightarrow e{{dy} \over {dx}} + 1 = 0 \Rightarrow {{dy} \over {dx}} = - {1 \over e}\]
\[ \Rightarrow {e^y}{{{d^2}y} \over {d{x^2}}} + {e^y}{\left( {{{dy} \over {dx}}} \right)^2} + x{{{d^2}y} \over {d{x^2}}} + 2{{dy} \over {dx}} = 0\]
x = 0, y = 1
\[ \Rightarrow e{{{d^2}y} \over {d{x^2}}} + e{\left( { - {1 \over e}} \right)^2} + 0 + 2\left( { - {1 \over e}} \right) = 0\]
\[ \Rightarrow {{{d^2}y} \over {d{x^2}}} = {1 \over {{e^2}}}\]