I did calculate $f_{xy}(x,y) and f_{yx}(x,y)$ and equating them gives the expression
$$tan(xy)=- \dfrac {xy}{2}$$
and equating $f_x(x,y)=0$ and $f_y(x,y)=0$ gives $$xy=(2m+1)\dfrac{\pi}{2}$$
These are the expressions I got:
$$f_x(x,y) = y\cos(xy)$$
$$f_y(x,y) = x\cos(xy)$$
$$f_{xx}(x,y) = -y^2\sin(xy)$$
$$f_{yy}(x,y) = -x^2\sin(xy)$$
$$f_{xy}(x,y) = -2y\sin(xy)-xy^2\cos(xy)$$
$$f_{yy}(x,y) = -2x\sin(xy)-x^2y\cos(xy)$$
How do I move from here?
Best Answer
You do not need the tools of calculus here, as we have a pretty good idea about how the $\sin$ function looks like. Also, I am not sure why you need cross partial derivatives.
$\sin(xy)$ has maximum value of $1$ at every $n\in \mathbb{Z}$ $xy = \pi/2 + 2n\pi$. Then, whenever you have $y=\pi\frac{1+4n}{2x}$, you will have a maximum.