Limits and Continuity in Multi variable calculus. 
Checking continuity of $f(x,y)$ at $(0,0)$:
   $$
f(x,y)=\begin{cases}\dfrac{x^3+y^3}{x-y}\ \ ,x\neq y\\0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ,x=y
\end{cases}$$

Using polar coordinates $x=r\cos\theta$ and $y=r\sin\theta$,
$$\lim_{r\rightarrow 0}\dfrac{r^3(\cos^3\theta+\sin^3\theta)}{r(\cos\theta-\sin\theta)}=0$$
Hence,$$\lim_{(x,y)\rightarrow (0,0)}f(x,y)=f(0,0)=0$$
But its graph doesn't look like continuous near-complete $z$-axis.
Being very new to this I don't know what's going on here, please help.

 A: Notice that taking the limit as $r \to 0$, like you did, only gets near zero through paths that are straight lines. This is not enough to ensure differentiability and this example is to show why. As pointed out in the comments, we have that
$$
\frac{\cos^3 \theta + \sin^3 \theta}{\cos \theta - \sin \theta}
$$
is not bounded, when you approach zero through the line $y=x$. Although the function is defined to be zero on this line, you can still replicate this exploding behavior by choosing some path that approaches line $y=x$ sufficiently fast.
If you consider the path $x(t) = t$ and $y(t) = t - t^3$, then you have
$$
\lim_{t \to 0} \frac{x^3 + y^3}{x - y}
= \lim_{t \to 0} \frac{t^3 + (t - t^3)^3}{t^3}
= \lim_{t \to 0}\frac{2t^3 - 3t^5 + 3t^7 - t^9 }{t^3}
= 2.
$$
A: Your mistake is in falsely assuming that $g(\theta) = \dfrac{\cos^3\theta+\sin^3\theta}{\cos\theta-\sin\theta}$ is bounded.  
It has a discontinuity at $\theta = \pi/4$  so it is not bounded.
Hence you cannot conclude $ \lim\limits_{r\to 0}f(r) g(\theta)=0$
