I'm a bit confused with this.
The gradient of the function is $$\nabla f=\left(\frac{x\sin A}{A} \quad \frac{y\sin A}{A}\right)^T $$ where $A=\sqrt{x^2+y^2}.$
One seemingly obvious solution of $$\nabla f= \bf{0}$$ is $(x,y)=(0,0)$, but it isn't one (or is it?) because $\nabla f$ isn't defined at that point.
The original function exists at $(0,0)$ and when I plot it it seems to have a stationary point at $(0,0)$.
I tried to resolve it by considering the $\lim_{(x,y) \rightarrow (0,0)} \frac{\sin A}{A}=1$, but I'm not convinced as even though it would make $\nabla f \rightarrow \bf{0}$ , the limit doesn't exist as the function is not defined.
Can someone please help me resolve this?
The gradient you wrote is ok for $(x,y)\ne(0,0)$ because the square root is not differentiable at $(0,0)$, for $(0,0)$ you must use the definition: $$f_x(0,0)=\lim_{h \to 0} \frac{2-\cos|h|-1}{h}=0$$ And similarly for $f_y(0,0)=0$.
Since it is $1 \le f(x,y) \le 3$ with $f(0,0)=1$, it follows that $(0,0)$ is a point of minimum for your $f$. From the gradient, you can deduce that other stationary points are for $\sin \sqrt{x^2+y^2}=0 \iff \sqrt{x^2+y^2}=k\pi$ with $k \in \mathbb{N}$, that is the points on the circumference of center $(0,0)$ and radius $\sqrt{k\pi}$. You can study these points with the parametrization $\gamma(t)=(\sqrt{k\pi} \cos t,\sqrt{k\pi} \sin t)$ for $0 \le t \le 2\pi$.