Proving (or disproving) multivariable limit existence I've come across a problem I think is impossible, but I could be wrong. It goes 

Let $$f(x, y) = \begin{cases}0,& \text{ if }y = 0\\ \ \\ y + x\sin\tfrac1y,&\ \text{ otherwise}\end{cases}$$
  Show that the limits
  $\lim\limits_{(x y) \to (0,0)}$ and $\lim\limits_{ y \to 0}$ $\lim\limits_{ x \to 0}$ $f(x,y)$ exist, while $\lim\limits_{ x \to 0}$ $\lim\limits_{ y \to 0}$ $f(x,y)$ does not exist. 

In the original equation, if you let $y=0$, then $f(x,0) = 0 + x\sin(1/0)$, which isn't $0$ because it's undefined, right? So how can we even continue with the problem?
And, if we can continue, how can the first limit $\lim\limits_{x y \to (0,0)}$ exist if $xy$ can only map to $0$, not $(0,0)$?

As you wrote yourself, when $y=0$ you have $f(x,y)=0$. 
You have 
$$
\lim_{x\to0}f(x,y)=\lim_{x\to0}y+x\sin\tfrac1y=y.
$$
Then 
$$
\lim_{y\to0}\lim_{x\to0}f(x,y)=\lim_{y\to0}\lim_{x\to0}y+x\sin\tfrac1y=\lim_{y\to0}y=0.
$$
The double limit is the same: you have
$$
|y+x\sin\tfrac1y|\leq|y|+|x|\leq2\sqrt{x^2+y^2}\to0
$$
when $(x,y)\to0$.
Finally, $\lim_{y\to0}y+x\sin\tfrac1y$ doesn't exist because of $\tfrac1y$. 
