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
  $(x,y)=(0,0)$
  
  $$
\begin{equation}
f(x,y) = 
\begin{cases}
\frac{x^3+y^3}{x-y} &amp;x\neq y\\
0  &amp;x = y
\end{cases}
\end{equation}
$$


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.
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 $y=mx^{1/4}$.
If $n\in\mathbb N$, then$$f\left(\frac1n,\frac1n+\frac1{n^3}\right)=-1-\left(1+\frac1{n^2}\right)^3$$and therefore $\lim_{n\to\infty}f\left(\frac1n,\frac1n+\frac1{n^3}\right)=-2$.