What is $\lim_{(x,y)\to(0,0)} \frac{(x^3+y^3)}{(x^2-y^2)}$?

If you define $$\lim_{\langle x,y\rangle\to\langle a,b\rangle}f(x,y)\tag{1}$$ in such a way that it exists only when the function is defined in some open ball centred at $\langle a,b\rangle$, then what you wrote is correct. This is analogous to defining $$\lim_{x\to a}f(x)$$ only when $f(x)$ is defined in some open interval centred at $a$. However, just as we can talk about one-sided limits on the real line, it makes perfectly good sense to talk about $(1)$ whenever $f(x,y)$ is defined at points arbitrarily close to $\langle a,b\rangle$. In that case it’s understood that we look only at the limit along ‘paths’ within the domain of $f$. On that understanding $$\lim_{\langle x,y\rangle\to\langle 0,0\rangle}\frac{x^3+y^3}{x^2-y^2}$$ still does not exist, but for a more fundamental reason.

Suppose that you approach the origin along the curve $y=\sin x$. Then by l’Hospital’s rule you have

$$\begin{align*} \lim_{\langle x,y\rangle\to\langle 0,0\rangle}\frac{x^3+\sin^3x}{x^2-\sin^2x}&=\lim_{\langle x,y\rangle\to\langle 0,0\rangle}\frac{3x^2+3\sin^2x\cos x}{2x-2\sin x\cos x}\\\\ &=\lim_{\langle x,y\rangle\to\langle 0,0\rangle}\frac{3x^2+\frac32\sin2x\cos x}{2x-\sin2x}\\\\ &=\lim_{\langle x,y\rangle\to\langle 0,0\rangle}\frac{6x-\frac32\sin2x\sin x+3\cos2x\cos x}{2-2\cos2x} \end{align*}$$

which does not exist: the numerator approaches $3$ and the denominator, $0$. The problem is that this path, although it stays within the domain of the function, approaches the line $y=x$ so quickly as it approaches the origin that the denominator approaches $0$ much faster than the numerator, and therefore the function blows up as we approach the origin along this path.

Brian's answer is nice, so it got me thinking, what about approaching the point along other functions, $y=f(x)$, with $f(0)=0$. Let us plug that into the expression to get $\frac{x^3+f^3}{x^2-f^2}$. This will give $$\lim_{x \to 0} \frac{x^2-xf+f^2}{x-f}\;.$$

We will now perform a l’Hospital’s rule to get $$\lim_{x \to 0} \frac{2x-xf'-f+2ff'}{1-f'}\;.$$ If at this point we assume that $f'(0)=1$, we have another $0/0$. If $f'\neq 1$, we get that the limit is zero. So letus now hit this with another l’Hospital. This will give us $$\lim_{x \to 0} \frac{2-2f'-xf''+2f''f'+2(f')^2}{-f''}\;.$$

When evaluate at $x=0, f'=1$, we get $$\lim_{x \to 0} -\frac{2f''+2}{f''}\;.$$ And you can decide what you want the limit to be by picking a value for $f''$.

It doesn't matter that along the lines $y = \pm x$, the function is undefined ; you can still stay inside the domain of $f(x,y) = \frac{x^3 + y^3}{x^2-y^2}$ and try to compute the limit, under the constraint that $y \neq \pm x$. The limit could still be defined at $(0,0)$.

But there are cases where your trick doesn't work : think of $\frac{x^4-y^4}{x^2-y^2} = x^2 + y^2$, which is continuous at $(0,0)$.

Hope that helps,