Prove with the $\epsilon - \delta$ proof that $\lim_{(x, y) \rightarrow (a, a)}\frac{x^3 - y^3}{x^2 - y^2} = \frac{3a}{2}$ So that is the question. Here is my partial work:
Let $x = s + a$ and $y = t + a$.
Then $d(x, y) = \sqrt{s^2 + t^2}$. If $\sqrt{s^2 + t^2} < \delta$, then $s^2 + t^2 < \delta^2$.
So then
\begin{eqnarray}
|\frac{x^3 - y^3}{x^2 - y^2} - \frac{3a}{2} &=& |\frac{(x - y)(x^2 + xy + y^2)}{(x - y)(x + y)} - \frac{1,5ax + 1,5ay)}{x + y}|\\
 &=& |\frac{x^2 + xy + y^2 - 1,5ax - 1,5ay}{x + y}| \\
&=&
|\frac{(s + a)^2 + (s + a)(t + a) + (t + a)^2 - 1,5(s + a) - 1,5(t + a)}{s + t + 2a}| \\
&=&
|\frac{s^2 + 2as + a^2 + st + as + at + a^2 + t^2 + 2at + a^2 - 1,5as - 1,5at - 3a^2}{s + t + 2a}| \\
&=&
 |\frac{s^2 + t^2 + 1,5as + 1,5at + st}{s + t + 2a}|
\end{eqnarray}
Then I'm stuck. I also found that $|s + t + 2a| > |a|$ with the reverse triangle inequality, but that didn't really help because you're still stuck with the $st$ term.

Here is one approach: if $a=0,\ $ the result is almost immediate. Otherwise, assume $a\neq 0,\ $ set $x-a=\delta \cos t, y-a=\delta \sin t,\ $ 
take $0<\delta <1$ so small that $|2a+\delta (\sin t+\cos t)|>\frac{|a|}{2},\ $
Then, 
$\left | \frac{\left ( (a+\delta \cos t)^{2}+((a+\delta \cos t))(a+\delta \sin t)+(a+\delta \sin t)^{2} \right )}{2a+\delta (\cos t+\sin t)}-\frac{3a}{2} \right |=\\ \left | \frac{3a^{2}+\delta ^{2}+\delta ^{2}\cos t\sin t+2\delta a\cos t+\delta a(\sin t+\cos t)+2a\delta \sin t  }{2a+\delta (\cos t+\sin t)}-\frac{3a}{2} \right |=\\ \left | \frac{3a^{2}}{2a+\delta (\cos t+\sin t)}+\frac{\delta ^{2}+\delta ^{2}\cos t\sin t+2\delta a\cos t+\delta a(\sin t+\cos t)+2a\delta \sin t}{2a+\delta (\cos t+\sin t)}-\frac{3a}{2} \right |\le \\ \left | \frac{3a^{2}}{2a+\delta (\cos t+\sin t)}-\frac{3a}{2} \right |+\left | \frac{\delta ^{2}+\delta ^{2}\cos t\sin t+2\delta a\cos t+\delta a(\sin t+\cos t)+2a\delta \sin t}{2a+\delta (\cos t+\sin t)} \right |\le\\ \left | \frac{3a^{2}}{2a+\delta (\cos t+\sin t)}-\frac{3a}{2} \right |+\delta \left ( \frac{2+5|a|}{|a|/2} \right )$ 
and now each term can be handled easily. 
