The "iterated" limit
$$ \lim_{x\to 0} \lim_{y\to 0} x\sin \Bigl(\frac{1}{y}\Bigr)$$
does not exists, because it does not exist the inner limit.
EDIT: my first treatment of the limit for $(x,y)\to (0,0)$ was not correct. I think that now all works well.
We have to be careful about the following limit in $\mathbb{R}^2$
$$\lim_{(x,y)\to (0,0)} x\sin \Bigl(\frac{1}{y}\Bigr)$$
Call for simplicity $f(x,y) = x\sin \Bigl( \frac{1}{y}\Bigr)$.
The point $(0,0)$ is an accumulation point for pairs of the form $(x,0)$ where the inner function is not defined. It turns out that the limit does not exist, because the sequence $(\frac{1}{n},0)$ converges to $(0,0)$ as $n$ approaches $+\infty$, but $f(\frac{1}{n},0)$ does not converge to any value (since $f$ is not defined at every point of the sequence); while at the same time, there are infinitely many sequences $(x_n,y_n)$ converging to $(0,0)$ such that $f(x_n,y_n)$ converges to $0$.
However, it is meaningful to consider the limit
$$\lim_{\substack{(x,y)\to (0,0) \\ y \neq 0}} x\sin \Bigl(\frac{1}{y}\Bigr)$$
We are simply restricting ourselves to the domain of $f$. In this region, which is $D = \mathbb{R}^2 \setminus \{y=0\}$, the inner function is everywhere defined and we can argue in a simple way: the sine is bounded everywhere and the function $x$ is infinitesimal as $(x,y)\to (0,0)$. Since this estimate holds in the whole $D$, we conclude that the above limit exists and is equal to $0$.