Solución al ejercicio 6954

\frac{\partial ^2 f}{\partial x\partial x}=\frac{2(x^2+e^y)-4x^2}{(x^2+e^y)^2}
\frac{\partial ^2 f}{\partial x\partial y}=\frac{\partial ^2 f}{\partial y\partial x}=\frac{-2xe^y}{(x^2+e^y)^2}
\frac{\partial ^2 f}{\partial y\partial y}=\frac{e^y(x^2+e^y)-e^{2y}}{(x^2+e^y)^2}
H(f)=\left( \begin{array}{cc} 0 & -\frac{1}{2}  \\ -\frac{1}{2} & \frac{1}{4} \end{array} \right)