Answer
Local maximum: $f(\dfrac{\pi}{2},\dfrac{\pi}{2})=f(-\dfrac{\pi}{2},-\dfrac{\pi}{2})=1$
Local minimum $f(\dfrac{\pi}{2},-\dfrac{\pi}{2})=f(-\dfrac{\pi}{2},\dfrac{\pi}{2})=-1$
Saddle points are 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 $(x,y)=(0,0)$ $D(0,0)=4 \gt 0$ and $f_{xx}(0,0)=2 \gt 0$ For $(x,y)=(0,1)$ $D(0,1)=4 \gt 0$ and $f_{xx}=2 \gt 0$ For $(x,y)=(\pi,-1)$ $D(\pi,-1)=4 \gt 0$ and $f_{xx}=2 \gt 0$ For $(x,y)=(2\pi,1)$ $D(\pi,-1)=4 \gt 0$ and $f_{xx}=2 \gt 0$ For $(x,y)=(\dfrac{\pi}{2},0)$ $D(\dfrac{\pi}{2},0)=-2 \lt 0$ For $(x,y)=(\dfrac{3\pi}{2},0)$ $D(\dfrac{\pi}{2},0)=-2 \lt 0$
Also, $f(\dfrac{\pi}{2},\dfrac{\pi}{2})=\sin (\dfrac{\pi}{2}) \sin (\dfrac{\pi}{2})=1$
$f(-\dfrac{\pi}{2},-\dfrac{\pi}{2})=\sin (-\dfrac{\pi}{2}) \sin (-\dfrac{\pi}{2})=1$
$f(\dfrac{\pi}{2},-\dfrac{\pi}{2})=\sin (\dfrac{\pi}{2}) \sin (-\dfrac{\pi}{2})=-1$
$f(-\dfrac{\pi}{2},\dfrac{\pi}{2})=\sin (-\dfrac{\pi}{2}) \sin (\dfrac{\pi}{2})=-1$
Hence, Local maximum: $f(\dfrac{\pi}{2},\dfrac{\pi}{2})=f(-\dfrac{\pi}{2},-\dfrac{\pi}{2})=1$
Local minimum $f(\dfrac{\pi}{2},-\dfrac{\pi}{2})=f(-\dfrac{\pi}{2},\dfrac{\pi}{2})=-1$
Saddle points are at $(0,0)$
