Non-existence of $\lim \limits_{(x, y) \to (0,0)} \frac{x^3 + y^3}{x - y} $ 
How to show that $\lim \limits_{(x, y) \to (0,0)} f(x, y)$ does not
  exist where,   
$$f(x, y) =  \begin{cases} \dfrac{x^3 + y^3}{x - y} \; ; & x \neq y \\
 0 \; \;\;\;\;\;\;\;\;\;\;\; ; & x = y  \end{cases}  $$

I tried bounding the value of the function as $(x, y)$ approaches $(0,0)$ but was not successful. I graphed the function and saw that it was actually approaching $0$ but Microsoft Math failed to render some points around the $x = y$ plane. So I imagine I need to find points $(x, y)$ such that $x$ is extremely close to $y$ but $(x, y)$ is not as close to $(0,0)$ to show that the function does not approach $0$ around the origin. But was unable to think of any. 
Hints or suggestions would be awesome. Any help is appreciated.
 A: Hint:  What happens on the curves $$(x(t), y(t)) = (1/t \pm e^{-t}, 1/t), \quad t > 0?$$
Here is a little animation of the surface in question.  It was surprisingly challenging to obtain a smooth plot in the neighborhood of $(0,0)$, but I managed to find a nice parametrization.  The red and green curves are the ones I described, but it is quite clear that other paths to the origin will also work.

A: For all $(x,y)\in \mathbb R^2$ such that $x\neq y$ one has $f(x,y)=\dfrac{2x^3}{x-y}-x^2-xy-y^2$, so if the limit exists, due to $\lim \limits_{(x,y)\to(0,0)}\left(x^2-xy-y^2\right)$ existing, so does $\lim \limits_{(x,y)\to (0,0)}\left(\dfrac{2x^3}{x-y}\right)$, but this last limit can easily be seen to be $k$-dependent if $y=x-kx^3$.
So consider the paths $t\mapsto\left(t,t-kt^3\right)$, with $k\neq 0$.
A: You can show this limit does not exist if you can show that the limits as $(x,y) \rightarrow (0,0)$  along two different paths give different answers.
So for example, try the limit as $s\rightarrow 0$ of $f(x=s, y=-s)$, which is zero,
and the limit as $s \rightarrow 0$ of $f(x=s, y=\sqrt[3]{s})$ which is $-1$.
