What is the critical points of $f(x,y) = e^{\sin x\cos y} $?

Question

I try to find local extreme values and saddle point(s) of the $f(x,y) = e^{\sin x\cos y} $.

But, when I take the partial derivative, I can't figure out to find critical points.

$$f(x,y) = e^{\sin x\cos y} $$ $$f_x(x,y) = \cos x\cos y\, e^{\sin x\cos y} = 0 $$ $$f_y(x,y) = -\sin x\sin y\,e^{\sin x\cos y} = 0 $$

How does that work?

Thanks in advance.

Answer 1

$$\cos x \cos y =0$$ $$\sin x \sin y =0$$

So $\cos x = \sin y =0$ or $\sin x = \cos y = 0 $

So $x=\frac{\pi}{2} + n\pi$ and $y=m\pi$

Or $y=\frac{\pi}{2} + l \pi$ and $x=k\pi$

These are points $(\frac{\pi}{2} + n\pi, m\pi)$ and $(k\pi, \frac{\pi}{2} + l \pi)$ with $k,l,m,n \in \mathbb Z$

Answer 2

Hint: You need $\cos x \cos y=0$, and also $\sin x \sin y=0$. However, $\cos x$ and $\sin x$ are never zero at the same time.