Show that $f(x,y) = \frac{x^3+y^3}{x-y}$ when $x \neq y$ is discontinuous at the origin 
Show that the following function is discontinuous at the origin
  $(x,y)=(0,0)$
$$
\begin{equation}
f(x,y) = 
\begin{cases}
\frac{x^3+y^3}{x-y} &x\neq y\\
0  &x = y
\end{cases}
\end{equation}
$$

Now I am taking $x=my\ (m \neq 1)$ and getting $f(x,y)$ as 
$$\lim_{(x,y) \to (0,0)}f(x,y)=\lim_{x \to 0}\frac{x^3+m^3x^3}{x-mx} = \lim_{x \to 0}\frac{x^2+m^3x^2}{1-m}$$
which tends to zero showing that the function is continuous. But we need to show discontinuance, I am not getting any other substitution for $y$ that will prove discontinuity.
 A: 
which tends to zero showing that the function is continuous 

No, the existence of the limit of $f(x,y)$ as $(x,y)$ approaches $(0,0)$ along one path does not show that the function is continuous at $(0,0)$. To prove continuity, you need to prove this limit exists (and has the same value) along every path to $(0,0)$. 
Thus to disprove continuity, you only need to show the non-existence of the limit, or the existence of a limit with a value different from $f(0,0)$, along a single path to $(0,0)$. You already tested all linear paths, so try something that's not linear. 
It might be helpful to use long division to write:
$$\frac{x^3+y^3}{x-y} = x^2+xy+y^2+\frac{2y^3}{x-y}, \quad x \neq y.$$
The $x^2+xy+y^2$ portion is continuous at $(0,0)$, so the $\frac{2y^3}{x-y}$ portion is the discontinuous portion. Notice that the numerator is cubic in $y$, while the denominator is linear in $x$ and $y$, so (as you found) if $y$ is a linear function of $x$ then the numerator will approach $0$ faster than the denominator. To get a limit other than $0$, you want the numerator to approach zero no faster than the denominator, which suggests trying the path $y=mx^{1/3}$, or something that goes even slower like $y=mx^{1/4}$.
A: 
(...) which tends to zero showing that the function is continuous.

No, that's not something you can conclude from these limits only. Limits along all paths have to be equal for the function to be continuous and that's not the case. 
Finding a path which results in a non-zero limit will be sufficient to conclude that the function is not continous in the origin.
A: Along the path $y=x-x^3$, the limit approaches to $2$.
