What is $\lim_{(x,y) \to (0,0)} \arctan(xy)/\sqrt{x^2+y^2}$?

Observe that you have $$ \left|\arctan u\right|\leq|u|, \quad |u|\leq1, $$ then, switching to polar coordinates with $r=\sqrt{x^2+y^2}$, as $r\to 0$, you get

$$ \left|\frac{\arctan(xy)}{\sqrt{x^2+y^2}}\right|=\frac{\left|\arctan(r^2 \sin \theta \cos \theta)\right|}{r}\leq \frac{\left|r^2 \sin \theta \cos \theta\right|}{r}\leq r. $$

The sought limit is equal to $0$.

$-\frac{x^2+y^2}{2}\leq xy\leq \frac{x^2+y^2}{2}$ and $\arctan$ is a monotonic increasing function, so $$\frac{\arctan\frac{-r^2}{2}}{r}\leq \frac{\arctan{(xy)}}{\sqrt{x^2+y^2}}\leq \frac{\arctan\frac{r^2}{2}}{r}$$ where $r=\sqrt{x^2+y^2}$. Now, using L'Hospital's rule you get that both limits in the sandwich are $0$.

Or, you can use the inequality Oliver Oloa gave to get: $$-\frac{r^2}{2r}\leq\frac{\arctan\frac{-r^2}{2}}{r}\leq \frac{\arctan{(xy)}}{\sqrt{x^2+y^2}}\leq \frac{\arctan\frac{r^2}{2}}{r}\leq\frac{r^2}{2r}$$

Use polar coordinates and L'Hospital's rule:

$$\lim\limits_{(x,y) \to (0,0)} \frac{\arctan(xy)}{\sqrt{x^2+y^2}} = \lim\limits_{r \to 0} \frac{\arctan(r^2 \cos(\theta) \sin(\theta))}{\sqrt{(r \cos(\theta))^2+(r \sin(\theta))^2}} = \lim\limits_{r \to 0} \frac{\arctan(r^2 \cos(\theta) \sin(\theta))}{r} = \lim\limits_{r \to 0} \frac{2r \cos(\theta) \sin(\theta))}{r^2 \sin^2(\theta)\cos^2(\theta) + 1} = 0$$