$f(x,y)={ e }^{ xy }\sin(y)\\ R=\left\{ (x,y):\left| x \right| \le 1;\left| y \right| \le 1 \right\} $
The partial Derivatives are: $f^{ x }\left( x,y \right) =y{ e }^{ xy }\sin(y)\\ f^{ y }\left( x,y \right) =(x\sin(y)+\cos(y)){ e }^{ xy }$
After setting both of them to zero, we get the requirement that $y=0$ or $y=k\pi$. But those Points dont statisfie the second equation,in result showing us there are no Critical Points in the inside of $R$. Since we know the set is compact Max/Min need to exist, so we need to evaluate the boundaries. Here I'm Stuck. Since the boundary is $\left| x \right| =1\quad \left| y \right| =1$ we would take those values and plug it into our function thus being only dependend on one Variable but here I get functions here i can't get a Max/min off. Show how would I procced from here?
If $x\in[-1,1]$, then $f(x,-1)=e^{-x}\sin(-1)=-\frac{\sin 1}{e^x}$. Therefore, in $[-1,1]\times\{-1\}$ the maximum is $-\frac{\sin1}e$ and the minimum is $-e\sin1$. Now, do the same thing in $[-1,1]\times\{1\}$, in $\{-1\}\times[-1,1]$ and in $\{1\}\times[-1,1]$.