I'll try to be more elaborate on the point I was trying to get across on the IRC earlier today.
Let's consider the following function, defined in the polar coordinates: $f: [0, \infty) \times (0, 2\pi] \to \mathbb{R}$,
$$
f(r, \theta) = \frac{r}{\theta}
$$
This is well defined, because our angle argument, $\theta$, varies in $(0, 2\pi]$, and not in $[0, 2\pi)$ as usual. Obviously this does not matter at all, because $2\pi$ determines the same angle as $0$. This function determines a function $\mathbb{R}^2 \to \mathbb{R}$ in a natural way, and I will denote that function by $f$ as well, using $x, y$ coordinates instead of $r, \theta$, hoping that will not cause any confusion. Also, $f(0) = 0$ (because the point $(0, 0) \in \mathbb{R}^2$ is represented in polar coordinates as $(0, \theta)$ for any choice of $\theta$, and $f(0, \theta) = \frac{0}{\theta} = 0$).
So, what's the limit:
$$
\lim_{(x, y) \to (0, 0)} f(x, y) = \;?
$$
Let's try to solve it using your approach:
$$
\lim_{(x, y) \to (0, 0)} f(x, y) = \lim_{r \to 0} \frac{r}{\theta} = 0
$$
Easy, wasn't it? And the resulting value does not depend on $\theta$. Let's conclude thus that the limit is $0$ and call it a day.
Right? Wrong! Consider a sequence (in polar coordinates):
$$
x_n = \left(r_n, \theta_n\right) = \left(\frac{1}{n}, \frac{\pi}{n}\right)
$$
Basically, we start with $x_1$ which is $(-1, 0)$ in cartesian coordinates, and we go clockwise, approaching half a turn, and going closer and closer to $(0, 0)$ in process. It's obvious that as a sequence of point in a plane, this sequence tends to $(0, 0)$ (because, well, in cartesian coordinates it's $x_n = (r_n \cos \theta_n, r_n \sin \theta_n) = (\frac{1}{n} \cos(\frac{\pi}{n}), \frac{1}{n} \sin(\frac{\pi}{n}))$, and obviously each component tends to $0$ as $n \to \infty$).
Now, if $\lim_{(x, y) \to (0, 0)} f(x, y)$ was in fact $0$, then we would also have $\lim f(x_n) = 0$, because $x_n \to (0, 0)$. But, surprisingly:
$$
\lim f(x_n) = \lim f(r_n, \theta_n) = \frac{\frac{1}{n}}{\frac{\pi}{n}} = \frac{1}{\pi} \ne 0
$$
What happened? It turns out, that $\lim_{r \to 0} f(r, \theta) = 0$ does not imply that $\lim_{(x, y) \to (0, 0)} f(x, y) = 0$. When we consider the limit as $r \to 0$, we keep $\theta$ fixed, so we only take a limit along a single straight line passing through the origin. This only implies that if we restrict the our function to any such straight line $L$, and consider a function $f|_L: L \to \mathbb{R}$ it will in fact be continuous at $0$. The problem is that the fact that the restricted function $f|_L: L \to \mathbb{R}$ is continuous at origin for every $L$ passing through origin, it's not enough to conclude that our function is in fact continuous at origin.
No need to go back to the definition of limits, use some of the properties instead. If $f(x,y)$ is continuous at $(0,0)$, consider $p(x) = (x, 0)$; $p$ is continuous everywhere (perhaps you need to prove that, but that's easy). So: if $f$ is continuous at $(0,0)$, then $g(x) = f \circ p (x)$ is continuous at $0$. Compute $f \circ p$ and show it is not continuous at $0$ for either $a=0$ or $a=1$, again, this is much easier than the original problem. So the conclusion is that the original $f$ is not continuous at $(0,0)$ for either value of $a$.
I gave the argument in relatively formal terms; but what I am doing is this: If $f$ is continuous as a function of two variables, it should in particular be continuous when I fix $y=0$ - and it just isn't.
Best Answer
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 $$ \lim_{t \to 0} \frac{x^3 + y^3}{x - y} = \lim_{t \to 0} \frac{t^3 + (t - t^3)^3}{t^3} = \lim_{t \to 0}\frac{2t^3 - 3t^5 + 3t^7 - t^9 }{t^3} = 2. $$