What does $\lim\limits_{(x,y)\rightarrow0}$ mean and how to show $ \lim\limits_{(x,y)\rightarrow0}\frac{x^3}{x^2+y^2}=0$? Consider $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ where
$$f(x,y):=\begin{cases}
      \frac{x^3}{x^2+y^2} & \textit{ if }  (x,y)\neq (0,0) \\
      0 & \textit{ if }  (x,y)= (0,0)
    \end{cases} $$
If one wants to show the continuity of $f$, I mainly want to show that 
$$ \lim\limits_{(x,y)\rightarrow0}\frac{x^3}{x^2+y^2}=0$$
But what does $\lim\limits_{(x,y)\rightarrow0}$ mean? Is it equal to $\lim\limits_{(x,y)\rightarrow0}=\lim\limits_{||(x,y)||\rightarrow0}$ or does it mean $\lim\limits_{x\rightarrow0}\lim\limits_{y\rightarrow0}$?
If so, how does one show that the above function tends to zero?

The formal definition is as follows: given a function of $n$ real variables (here $n=2$): $f(x_1,\ldots, x_n),$ we say that 
$$\lim_{(x_1,\ldots, x_n)\to (p_1,\ldots, p_n)}f(x_1,\ldots,x_n)=L$$
if for every $\epsilon>0$, there exists a $\delta$ sufficiently small that $$ \lvert (x_1,\ldots, x_n)-(p_1,\ldots, p_n)\rvert<\delta$$ 
implies that 
$$ \lvert f(x_1,\ldots, x_n)-L\rvert<\epsilon.$$
In your case, this reduces to showing that for every $\epsilon>0$, there exists a $\delta$ sufficiently small that 
$$ \lvert (x,y)\rvert<\delta$$
implies that 
$$ \lvert f(x,y)\rvert<\epsilon.$$
Once you've digested this definition, it is worthwhile to observe that as $(x,y)\to 0$, we have that 
$$ \bigg|\frac{x^3}{x^2+y^2}\bigg|\le \lvert x\rvert\to 0.$$
