Disprove or prove using delta-epsilon definition of limit that $\lim_{(x,y) \to (0,0)}{\frac{x^3-y^3}{x^2-y^2}} = 0$ I want to prove if the following limit exists, using epsilon-delta definition, or prove it doesn't exist:$$\lim_{(x,y) \to (0,0)}{\frac{x^3-y^3}{x^2-y^2}} = 0$$
My attempt: 
First I proved some directional limits, like for $y=mx$ , and $y=ax^n$, and for all of them I got 0. So I conjectured that this limit exists and it's 0. Then I have to prove:
$$\forall \delta \gt 0 : \exists \epsilon \gt 0 : \|(x,y)\| \lt \epsilon \rightarrow \left| \frac{x^3-y^3}{x^2-y^2}\right| \lt \delta$$
First I noted that $\frac{x^3-y^3}{x^2-y^2} = \frac{(x^2+xy+y^2)(x-y)}{(x+y)(x-y)} = \frac{x^2+xy+y^2}{x+y}$.
Then I did $\left|\frac{x^2+xy+y^2}{x+y}\right| \leq \left|\frac{x(x+y)}{x+y}\right|+\frac{y^2}{\vert x+y\vert} = \vert x \vert + \frac{y^2}{\vert x+y\vert}$
Using $\|(x,y)\| = \vert x\vert + \vert y\vert$ and assuming  $\|(x,y)\| \lt \epsilon$
$\vert x\vert + \vert y\vert  = \vert x \vert + \frac{ y^2}{\vert y \vert} \geq \vert x \vert+\frac{y^2}{\vert y\vert+\vert x\vert}$
but I can't continue from that since $\vert x + y \vert \leq \vert x \vert + \vert y \vert$
.
I don't know what else to try.

OK, here is a completely different answer using $(x,y)$ coordinates only and no parameters.
Basically, try a power curve $y=x^n$.
BUT this gives you a path approaching the origin near the $x$ or $y$ axis, which is not a problem.  So we need to modify it to place it near the "problem" line $y=-x$.
Consider the curve
$$y=-x+x^2\ .$$
This will lie within the domain of your function as long as $x\ne0,2$.  So we can take $x\to0$, which implies $y\to0$, so the curve approaches the origin.  Along this curve we have
$$\frac{x^3-y^3}{x^2-y^2}=\frac{2x^3-3x^4+3x^5-x^6}{2x^3-x^4}
  =\frac{2-3x+3x^2-x^3}{2-x}$$
which tends to $1$ as $x\to0$.  Hence the limit does not exist.
