Answer
Maximum value:$f(\dfrac{1}{\sqrt 2},\dfrac{1}{\sqrt 2})=f(-\dfrac{1}{\sqrt 2},-\dfrac{1}{\sqrt 2})=0$
Minimum value: $f(-\dfrac{1}{\sqrt 2},\dfrac{1}{\sqrt 2})=f(\dfrac{1}{\sqrt 2},-\dfrac{1}{\sqrt 2})=-\dfrac{\sqrt2}{e}$
Saddle point at $(0,0)$
Work Step by Step
Second derivative test: Some noteworthy points to calculate the local minimum, local maximum and saddle point of $f$.
1. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum.
2.If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\lt 0$ , then $f(p,q)$ is a local maximum.
3. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ , then $f(p,q)$ is a not a local minimum and local maximum or, a saddle point.
For $(x,y)=(\dfrac{1}{\sqrt 2},\dfrac{1}{\sqrt 2})$ and $f(-\dfrac{1}{\sqrt 2},-\dfrac{1}{\sqrt 2})$
$D=2e^{-1} \gt 0$ ; and $f_{xx} =-2e^{-1}\lt 0$
$D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\lt 0$ , then $f(p,q)$ is a local maximum.
For $(x,y)=(\dfrac{1}{\sqrt 2},\dfrac{1}{\sqrt 2})$ and $f(\dfrac{1}{\sqrt 2},-\dfrac{1}{\sqrt 2})$
$D=2e^{-1} \gt 0$ ; and $f_{xx} =2e^{-1}\gt 0$
Thus, when $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum.
Therefore, we have Maximum value:$f(\dfrac{1}{\sqrt 2},\dfrac{1}{\sqrt 2})=f(-\dfrac{1}{\sqrt 2},-\dfrac{1}{\sqrt 2})=0$
Minimum value: $f(-\dfrac{1}{\sqrt 2},\dfrac{1}{\sqrt 2})=f(\dfrac{1}{\sqrt 2},-\dfrac{1}{\sqrt 2})=-\dfrac{\sqrt2}{e}$
Saddle point at $(0,0)$
