Answer
Local maximum: $f(\dfrac{\pi}{6},\dfrac{\pi}{6})=\dfrac{3}{2}$
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 $(x,y)=(\dfrac{\pi}{6},\dfrac{\pi}{6})$
$D=\dfrac{3}{4}\gt 0$ ; and $f_{xx} =-1\lt 0$
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,
$f(x,y)=sin x+sin y +cos(x+y)=sin x+ siny+cosx \cos y-\sin x \sin y$
This yields,
$f(\dfrac{\pi}{6},\dfrac{\pi}{6})=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{\sqrt 3}{2}\dfrac{\sqrt 3}{2}-\dfrac{1}{2}\dfrac{1}{2}=\dfrac{3}{2}$
Therefore, we have Maximum value: $f(\dfrac{\pi}{6},\dfrac{\pi}{6})=\dfrac{3}{2}$
