Finding extrema of $\frac{\sin (x) \sin (y)}{x y}$ I need to find the extrema of the following function in the range $-2\pi$ and $2\pi$ for both $x$ and $y$, but I don't know how to go about doing it since it's a bit weird and not similar to other functions I've seen:
$$f(x,y)=\frac{\sin (x) \sin (y)}{xy}$$
I've evaluated the gradient function as below:
$\nabla f = <\frac{\sin (y) (x \cos (x)-\sin (x))}{x^2 y}, \frac{\sin (x) (y \cos (y)-\sin (y))}{y^2 x}>$
but setting it to zero gives a few answers for $x$ and $y$, none of which seem to be the right answer.
The following is the sketch of the graph in the aforementioned range. According to WolframAlpha, it should have it's local extrema at $\{0, 4.49\}$, $\{0, -4.49\}$, $\{4.49, 0\}$, $\{-4.49, 0\}$.


As said in comments, if  $$F=\frac{\sin (x) \sin (y)}{xy}$$ $$F'_x=\frac{\cos (x) \sin (y)}{x y}-\frac{\sin (x) \sin (y)}{x^2 y}=\frac{\sin (y) (x \cos (x)-\sin (x))}{x^2 y}$$ $$F'_y=\frac{\sin (x) \cos (y)}{x y}-\frac{\sin (x) \sin (y)}{x y^2}=\frac{\sin (x) (y \cos (y)-\sin (y))}{x y^2}$$ where we see appearing the solutions of equations $z=\tan(z)$; beside the trivial solution $z=0$, there is only one solution which does not show any closed form. This solution is close to $\frac {3\pi}2$.
Developing $z-\tan(z)$ around $z=\frac {3\pi}2$ as a series, we have $$z-\tan(z)=\frac{1}{z-\frac{3 \pi }{2}}+\frac{3 \pi }{2}+\frac{2}{3} \left(z-\frac{3 \pi
   }{2}\right)+O\left(\left(z-\frac{3 \pi }{2}\right)^2\right)$$ and the positive root is given by $$z=\frac{1}{8} \left(3 \pi +\sqrt{81 \pi ^2-96}\right)\approx 4.49340$$ which is extremely close to the solution $(\approx 4.49341)$.
Then the solutions shown by Wolfram Alpha.
