$\begin{align}
  & {f}'(x)=1{{e}^{-{{x}^{2}}}}+(-2x){{e}^{-{{x}^{2}}}}.x={{e}^{-{{x}^{2}}}}(1-2{{x}^{2}}) \\
 & {f}''(x)=-2x{{e}^{-{{x}^{2}}}}(1-2{{x}^{2}})+(-4x){{e}^{-{{x}^{2}}}} \\
 & ={{e}^{-{{x}^{2}}}}(-2x+4{{x}^{3}}-4x) \\
 & {f}''(x)={{e}^{-{{x}^{2}}}}(4{{x}^{3}}-6x)=0\xrightarrow{{{e}^{-{{x}^{2}}}}\ne 0}4{{x}^{3}}-6x=0 \\
 & 2x(2{{x}^{2}}-3)=0\Rightarrow \left\{ \begin{matrix}
   {{x}_{A}}=0\Rightarrow {{y}_{A}}=k  \\
   {{x}_{B}}=\frac{\sqrt{6}}{2}\Rightarrow {{y}_{B}}=\frac{\sqrt{6}}{2}{{e}^{\frac{3}{2}}}+k  \\
   {{x}_{C}}=-\frac{\sqrt{6}}{2}\Rightarrow {{y}_{C}}=-\frac{\sqrt{6}}{2}{{e}^{-\frac{3}{2}}}+k  \\
\end{matrix} \right. \\
 & {{y}_{A}}+{{y}_{B}}+{{y}_{C}}=k+\frac{\sqrt{6}}{2}{{e}^{-\frac{3}{2}}}+k-\frac{\sqrt{6}}{2}{{e}^{-\frac{3}{2}}}+k \\
 & =3k=1\Rightarrow k=\frac{1}{3} \\
\end{align}$