Hallo,
Wolfram alpha bestätigt die Richtigkeit der Ergebnisse:
stationary points \( \quad\left(x^{2}+2 y^{2}\right) e^{-x^{2}-y^{2}} \)
\( \begin{array}{l}\left(x^{2}+2 y^{2}\right) e^{-x^{2}-y^{2}}=\frac{1}{e} \text { at }(x, y)=(-1,0) \quad \text { (saddle point) } \\ \left(x^{2}+2 y^{2}\right) e^{-x^{2}-y^{2}}=0 \text { at }(x, y)=(0,0) \quad \text { (minimum) } \\ \left(x^{2}+2 y^{2}\right) e^{-x^{2}-y^{2}}=\frac{2}{e} \text { at }(x, y)=(0,-1) \quad \text { (maximum) } \\ \left(x^{2}+2 y^{2}\right) e^{-x^{2}-y^{2}}=\frac{2}{e} \text { at }(x, y)=(0,1) \quad \text { (maximum) } \\ \left(x^{2}+2 y^{2}\right) e^{-x^{2}-y^{2}}=\frac{1}{e} \text { at }(x, y)=(1,0) \quad \text { (saddle point) }\end{array} \)