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$?

Note that we have

$$\left|\frac{x^3}{x^2+y^2}\right|\le |x|$$

The limit as $(x,y)\to(0,0)$ is therefore $0$.

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The limit $\lim_{(x,y)\to(0,0)}f(x,y)=L$ means that for all $\epsilon>0$, there exists a deleted neighborhood $N_{0,0}$ (e.g., there exists a $\delta>0$, such that $0<\sqrt{x^2+y^2}<\delta$), such that whenever $(x,y)\in N_{0,0}$, $|f(x,y)-L|<\epsilon$.

Note that the iterated limits $\lim_{x\to0}\lim_{y\to0}f(x,y$ and $\lim_{y\to0}\lim_{x\to0}f(x,y$ are not necessarily equal to each other or equal to the limit $\lim_{(x,y)\to(0,0)}f(x,y)$.

In THIS ANSWER, I referenced the Moore-Osgood Theorem, which gives sufficient conditions when the limit and the iterated limits are equal.

$\lim_\limits{(x,y)\to 0}$ likely means $\lim_\limits{(x,y)\to(0,0)}$, which means that $x$ and $y$ are both tending to $0$. One could use polar coordinates where $x=r\cos(\theta)$ and $y=r\sin(\theta)$ to obtain: $$\lim_{(x,y)\to(0,0)}\frac{x^3}{x^2+y^2}=\lim_{r\to 0}\frac{r^3\cos^3(\theta)}{r^2}=\lim_{r\to 0} r\cos^3(\theta)$$ Then note that $|\cos^3(\theta)|\leq 1~~~\forall\theta\in\Bbb R$.

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 $$ \bigg|\frac{x^3}{x^2+y^2}\bigg|\le \lvert x\rvert\to 0.$$