Hello Robin,
We are given the function:
$$f(x,y)=\sin(x)\sin(y)$$
where:
$$-\pi<x<\pi$$
$$-\pi<y<\pi$$
Let's take a look at a plot of the function on the given domain:
View attachment 1676
Equating the first partials to zero, we obtain:
$$f_x(x,y)=\cos(x)\sin(y)=0\implies x=\pm\frac{\pi}{2},\,y=0$$
$$f_y(x,y)=\sin(x)\cos(y)=0\implies x=0,\,y=\pm\frac{\pi}{2}$$
Adding, we find:
$$\sin(x)\cos(y)+\cos(x)\sin(y)=0$$
Applying the angle-sum identity for sine, we find:
$$\sin(x+y)=0$$
Observing that we require:
$$-2\pi<x+y<2\pi$$
We then have:
$$x+y=-\pi,\,0,\,\pi$$
Thus, we obtain the 5 critical points:
$$P_1(x,y)=\left(-\frac{\pi}{2},-\frac{\pi}{2} \right)$$
$$P_2(x,y)=\left(-\frac{\pi}{2},\frac{\pi}{2} \right)$$
$$P_3(x,y)=(0,0)$$
$$P_4(x,y)=\left(\frac{\pi}{2},-\frac{\pi}{2} \right)$$
$$P_5(x,y)=\left(\frac{\pi}{2},\frac{\pi}{2} \right)$$
To categorize these critical points, we may utilize the second partials test for relative extrema:
$$f_{xx}(x,y)=-\sin(x)\sin(y)$$
$$f_{yy}(x,y)=-\sin(x)\sin(y)$$
$$f_{xy}(x,y)=\cos(x)\cos(y)$$
Hence:
$$D(x,y)=\sin^2(x)\sin^2(y)-\cos^2(x)\cos^2(y)$$
[TABLE="class: grid, width: 660"]
[TR]
[TD]Critical point $(a,b)$[/TD]
[TD]$D(a,b)$[/TD]
[TD]$f_{xx}(a,b)$[/TD]
[TD]Conclusion[/TD]
[/TR]
[TR]
[TD]$\left(-\dfrac{\pi}{2},-\dfrac{\pi}{2} \right)$[/TD]
[TD]1[/TD]
[TD]-1[/TD]
[TD]relative maximum[/TD]
[/TR]
[TR]
[TD]$\left(-\dfrac{\pi}{2},\dfrac{\pi}{2} \right)$[/TD]
[TD]1[/TD]
[TD]1[/TD]
[TD]relative minimum[/TD]
[/TR]
[TR]
[TD]$(0,0)$[/TD]
[TD]-1[/TD]
[TD]0[/TD]
[TD]saddle point[/TD]
[/TR]
[TR]
[TD]$\left(\dfrac{\pi}{2},-\dfrac{\pi}{2} \right)$[/TD]
[TD]1[/TD]
[TD]1[/TD]
[TD]relative minimum[/TD]
[/TR]
[TR]
[TD]$\left(\dfrac{\pi}{2},\dfrac{\pi}{2} \right)$[/TD]
[TD]1[/TD]
[TD]-1[/TD]
[TD]relative maximum[/TD]
[/TR]
[/TABLE]
