Find the extrema and saddle points of $f (x,y)=e^x \sin(y)$ 
Find the extrema and saddle points of $f$
$f(x,y)=e^x \sin(y)$

My attempt:
$f_x=e^x \sin(y)$
$f_y=e^x \cos(y)$
Since $f_x$ and $f_y$ exist for all $(x,y)$ then the only critical points are the solutions of the following system:
$e^x \sin(y)=0$
$e^x \cos(y)=0$
But I can’t solve this system, can you help please?
Thanks.
 A: The function $e^x$ is never zero, so you can divide by it. Thus, you are looking for a solution of $\sin(y) = 0 = \cos(y)$. When, if ever, does this happen?
A: You are quite close to the answer:
Since you've already deduced that the critical points are where the following equations hold:
$$
\begin{align}
e^x sin(y) &= 0\\
e^x cos(y) &= 0
\end{align}
$$
Thus all that's left to do is to solve it. Consider what happens when we set $f_x$ to $0$:
$$
f_x = e^x sin(y) = 0
$$
Thus, $f_x$ takes on the value of zero when either $e^x = 0$ (not possible for any value of x, since $e^x > 0 \space \forall x$), or $sin(y)= 0$. Hence, we require $sin(y) = 0$. For which values of $y$ does this hold for? Well, it holds for:
$$
y = k\pi, \quad k\in \mathbb{Z}
$$
Applying the same treatment with $f_y$, we have $f_y = 0$ when $cos(y) = 0$, i.e. 
$$
y = (k +\frac{1}{2})\pi
$$
Now, what this means geometrically is that there will not be any point $(x,y)$ in which the two equations
$$
f_x = 0\\
f_y = 0
$$
hold, since $f_x \neq 0$ when $f_y = 0$, and $f_y \neq 0$ when $f_x = 0$. 
Thus, there are no saddle points nor any critical points.
