What is $\lim_{(x,y)\rightarrow(0,0)} \frac{x^3-y^3}{x^2-y^2}$? Given is the function $f(x,y)=\frac{x^3-y^3}{x^2-y^2}$ defined for $(x,y)\in\mathbb{R^2}, \, x^2\neq y^2$
I need to find the limit of $\lim_{(x,y)\rightarrow(0,0)} \frac{x^3-y^3}{x^2-y^2}$. 
I have tried a few paths, $(x,mx),\, (x,mx^2), \, (x,sin(x)), \, (x,tan(x))$ but all of them yield the same result, $0$. 
So I tried to prove the limit exists using the epsilon-delta definition. So:
$$\left|\frac{x^3-y^3}{x^2-y^2} - 0\right| = \left|\frac{(x-y)(x^2+y^2+xy)}{(x-y)(x+y)}\right|= \frac{x^2+y^2+xy}{\left|x+y\right|}<\epsilon$$
Here I am stuck and do not know how to proceed.

The problem is that there is no limit, only pathwise limits (which of course depend on the path to zero).
To see this one could rewrite the expression:
$${x^3-y^3\over x^2-y^2} = {(x-y)(x^2+xy+y^2)\over(x-y)(x+y)} = {(x+y)x - y^2\over (x+y)} = x - {y^2\over x+y}$$
Now of course $x\to 0$, but the term $y^2/(x+y)$ can be selected arbitrarily by having $x = y^2/C - y$. So along the parabola $x = y^2/C - y$ aproaching $0$ we have the limit of the expression being $-C$.  
But this can't happen if the limit actually exists.
