Is the domain of $f(x,y) = \dfrac{xy}{x^2 - y^2}$ an open set? And does the limit at $(0,0)$ exist? I am not all too sure if I get the whole concept of open and closed sets. 
So as an example I have this function:  $f(x,y) = \dfrac{xy}{x^2 - y^2}$.
I do understand that this is an open set because $G= \{(x,y) \in \Bbb R^2 : x \neq y \text{ and } x \neq -y\}$, meaning that every point of $G$ is an interior point. The boundaries are not included. Is this a correct way of thinking? 
Also the next question would be about the $\lim_{(x,y) \to (0,0)} f(x,y)$ not not existing. How can you show this? 
On a mini side note, the next set is closed because the boundaries are included in the set right? 
$$S = \{x \in \Bbb R^2 : x_1 ≥ 0, x_2 ≥ 0, x_1 + x_2 = 2\}. $$
I might be overthinking this but any help would be nice! 

One of the classical method that you can show that this kind of limits dont exist is this, you use $x=my^k$ or $y=mx^k$ where $m$ is variable and $k$ is integer. After substitution  $x=my^k$ or $y=mx^k$ in the limit, if the limit be dependent of the variable $m$, it means, it dosent limit. There are three note one) this method is acceptable in just in the case $(0,0)$. second) after substitution $x=my^k$ or $y=mx^k$, when you can say It dosent limit that in the limit dosent exist $x$ and $y$ and three) this method is necessary condition not sufficient, I mean with this method just you can show there is no limit not exist of the limit.For your question you should choose $y=mx$ and substitute it in the limit:
$$\frac{x(mx)}{x^2-(mx)^2}=\frac{mx^2}{x^2-m^2x^2}=\frac{m}{1-m^2}$$ 
we can see, the limit is dependent to $m$ and so there is no limit in $(0,0)$.
