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.
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 $y=mx^{1/4}$.