Answer
Local minimum of $f(0,0)=0$ and Saddle point at $(0,2)$
Work Step by Step
Since, we have $f_x(x,y)=2x e^ {-y}=0, f_y(x,y)=2ye^{-y}-e^{-y}(x^2+y^2)=0$
After solving the above two equations, we get
The critical points are: $(0, 0)$ and $(0,2)$
As per the second derivative test for $(0,0)$, we have
$D=f_{xx}f_{yy}-f^2_{xy}=4$ and $D=4 \gt 0$ and $f_{xx}=2 \gt 0$
Thus, we have: Local minimum of $f(0,0)=0$
As per the second derivative test, for $(0,2)$ we have
$D=f_{xx}f_{yy}-f^2_{xy}=-\dfrac{4}{e^4}$ and $D=-\dfrac{4}{e^4} \lt 0$
Thus, we have: Saddle point at $(0,2)$
Hence: Local minimum of $f(0,0)=0$ and Saddle point at $(0,2)$
