Exercise. Find and classify the critical points of $f(x,y)=\ln{(2+\sin{xy})}$.
I want to verify my work until now, and I also have a question about classifying the infinitely many critical points I'm getting. Here are the details:
To begin we compute $f_x(x,y)=\frac{y\cos{xy}}{\sin{xy}+2}$ and $f_y(x,y)=\frac{x\cos{xy}}{\sin{xy}+2}$. Then, to find the critical points, we solve the following system of equations:
$$\frac{y\cos{xy}}{\sin{xy}+2}=0$$ $$\frac{x\cos{xy}}{\sin{xy}+2}=0$$
which is equivalent to
$$y\cos{xy}=0\label{1}\tag{1}$$ $$x\cos{xy}=0\label{2}\tag{2}$$
From $\eqref{1}$ follows that $y=0$ or $\cos{xy}=0$.
If $y=0$, then $x=0$. Thus, (0,0) is a critical point.
If $\cos{xy}=0$, then $xy=\frac{\pi+4\pi{}n}{2} \Rightarrow x=\frac{\pi(4n+1)}{2y}$, for all $n\in\mathbb{Z}, y\ne{0}$. So, there are infinitely many critical points in the form $\big(\frac{\pi(4n+1)}{2y},y)$, for all $n\in\mathbb{Z}, y\ne{0}$.
Questions: Is this right? If so, how do I proceed to classify the critical points? (I know of the Hessian method, should I use it for all critical points with an analysis by cases?).
Maxima and minima of $\ln (2+\sin (xy))$ are same as those of $2+\sin (xy)$ because $\ln$ is strictly increasing. When $\cos (xy)=0$ either $\sin (xy)=1$ or $\sin (xy)=-1$. The former give local maxima and the latter give local minima.