Answer
Maximum value: $f(\dfrac{\pi}{3},\dfrac{\pi}{3})=\dfrac{3 \sqrt 3}{2}$
Minimum value: $f(\dfrac{5\pi}{3},\dfrac{5\pi}{3})=-\dfrac{3 \sqrt 3}{2}$
Saddle point at $(\pi, \pi)$
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 not a local minimum and local maximum or, a saddle point.
For $(x,y)=(\dfrac{\pi}{3},\dfrac{\pi}{3})$
$D=\dfrac{9}{4}\gt 0$ ; and $f_{xx} =\sqrt 3\gt 0$
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.
For $(x,y)=(\dfrac{5\pi}{3},\dfrac{5\pi}{3})$
$D=\dfrac{9}{4}\gt 0$ ; and $f_{xx} =-\sqrt 3\lt 0$
Thus, when $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\lt 0$ , then $f(p,q)$ is a local maximum.
Also, $D(\pi, \pi)=\sin x \sin y+\sin x \sin (x+y)+\sin y \sin (x+y)=0$
Therefore, we have Maximum value: $f(\dfrac{\pi}{3},\dfrac{\pi}{3})=\dfrac{3 \sqrt 3}{2}$
Minimum value: $f(\dfrac{5\pi}{3},\dfrac{5\pi}{3})=-\dfrac{3 \sqrt 3}{2}$
Saddle point at $(\pi, \pi)$
