Why is $f(x,y)$ discontinuous at $(x,y)=0$ 
$\displaystyle f(x,y)=\begin{cases} \displaystyle\frac{x^3+y^3}{x-y},
 x\ne y \\ 0, x=y\end{cases}$

The questions is to check continuity at $(x,y)=(0,0)$ for $f(x,y)$ 
My attempt: 
$f(x,y)$ is continuous at $(0,0)$ iff $\displaystyle \lim_{(x,y) \to(0,0)} f(x,y) = f(0,0)$ 
Now for $y=mx,m\ne 1$ 
$\displaystyle \lim_{{(x,y) \to(0,0)}_{y=mx}} f(x,y) = \lim_{{(x,y) \to(0,0)}_{y=mx}} \frac{x^3+y^3}{x-y} =\lim_{x \to 0} \frac{x^3+m^3x^3}{x-mx}= \lim_{x \to 0} x^2\left(\frac{1+m^3}{1-m}\right)=0$ 
Now for $y=x$ 
$\displaystyle \lim_{{(x,y) \to(0,0)}_{y=x}} f(x,y) = \lim_{{(x,y) \to(0,0)}_{y=x}} 0 = 0$ 
From the above we can say $\displaystyle \lim_{(x,y) \to(0,0)} f(x,y)=0$ 
$\implies \displaystyle \lim_{(x,y) \to(0,0)} f(x,y)=f(0,0)$ 
So that means $f(x,y)$ is continuous at $(0,0)$ 
But the source from which I got this question says it's discontinuous at $(0,0)$ 
I am not able to find my mistake 
Although I can verify limit using $\epsilon,\delta$ method but I couldn't 
So I need your help to prove me correct/wrong 
Also using $\epsilon,\delta$ method would be apreciated
 A: You have explored all of the obvious linear approaches to the point - however, the fact that the line is defined in a special way along $y=x$ is a hint that behaviour is strange near that line.
Consider the line $y=x-f(x)$, where $f(0)=0$. If we choose $f(x)$ such that $f'(0)=0$ as well, then in the neighbourhood of $(0,0)$, it will behave like $y=x$... but if $f(x)\neq0$ in this region, then it will not be on the line $y=x$.
If you restrict attention to this line, you are looking at
$$
\lim_{x\to0} \frac{x^3+(x-f(x))^3}{f(x)}=\lim_{x\to0}\ \left(\frac{2x^3}{f(x)}-3x^2+3xf(x)-f(x)^2\right) = \lim_{x\to0}\frac{2x^3}{f(x)}
$$
It's easy to see that $f(x)=cx^n$ will produce a nonzero result for $n\geq3$.
A: Along all the lines $y=mx$ with $m\not=1$ the limit is zero (you are correct), but this is not sufficient to conclude that the full limit is zero. Instead try the restriction to a curve through the origin which passes near the line $y=x$ which is not in the domain of $f$.
For instance, evaluate the limit along the curve $y=x+x^a$ with $a>1$ as $x\to 0^+$:
$$\frac{x^3+y^3}{x-y}=\frac{x^3+(x+x^a)^3}{-x^a}=\frac{2x^3+o(x^3)}{-x^a}\to \begin{cases} 0 &\text{if $1<a<3$}\\
-2 &\text{if $a=3$}\\
-\infty&\text{if $a>3$}\end{cases}.$$
Since the limit is different from zero for $a\geq 3$, the function is not continuous at $(0,0)$.
A: For $z\in {\mathbb R}$, let $\varphi(z) = (1+z)^3 + z^3$. This is a strictly increasing bijection from ${\mathbb R}$ onto ${\mathbb R}$. For $v\in {\mathbb R}$ and $t > 0$, let $z = \varphi^{-1}(v/t^2)$ and $x = t(1+z)$ and $y=tz$. Then
\begin{equation}
f(x, y) = \frac{t^3(1 + z)^3 + t^3 z^3}{t(1+z) - tz}  = 
\frac{t^3 \varphi(\varphi^{-1}(v/t^2))}{t} = t^2\frac{v}{t^2} = v
\end{equation}
when $t\to 0^+$, we have $z\sim \left(\frac{v}{2t^2}\right)^{1/3}$ hence $y\sim \left(\frac{v t}{2}\right)^{1/3}\to 0$ and $x = t + y\to 0$. It follows that any $v\in{\mathbb R}$ is a limit point of $f(x, y)$ when $(x, y)\to (0,0)$.
