Answer
Local minimums at $f(0,1)=f(\pi,-1)=f(2 \pi,1)=-1$ and saddle points are at $(\dfrac{\pi}{2},0),(\dfrac{3\pi}{2},0)$
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 not a local minimum, 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 $(x,y)=(\dfrac{3\pi}{2},0)$
$D(\dfrac{\pi}{2},0)=-2 \lt 0$
Hence,
Local minimums at $f(0,1)=f(\pi,-1)=f(2 \pi,1)=-1$ and saddle points are at $(\dfrac{\pi}{2},0),(\dfrac{3\pi}{2},0)$
