Show the Cauchy-Riemann equations hold but f is not differentiable 
Let $$f(z)={x^{4/3} y^{5/3}+i\,x^{5/3}y^{4/3}\over x^2+y^2}\text{ if }z\neq0
\text{, and }f(0)=0$$ 
  Show that the Cauchy-Riemann equations hold at $z=0$ but $f$ is not differentiable at $z=0$

Here's what I've done so far:
$\quad$As noted above, there are two cases: $z=0$ and $z\neq0$. The Cauchy-Riemann equations hold at $\quad z\neq0$ because $f(x)$ is analytic everywhere except at the origin.
$\quad$Now we consider $z=0$. In this case, 
$$f_x(0,0) = \lim_{x\to 0} \frac{0 + 0}{x^2+0}=0$$ 
$\quad$and
$$f_y(0,0) = \lim_{y\to 0} \frac{0 + 0}{0+y^2}=0$$ 
$\quad$So, we see that $\frac{\delta u}{\delta x}=0=\frac{\delta v}{\delta y}$ and $\frac{\delta u}{\delta y}=0= -\frac{\delta v}{\delta x}$. Therefore, the Cauchy-Riemann equations $\quad$hold at $z=0$. 
From here, I'm a little confused about how to show that $f$ is not differentiable at $z=0$. I instinctively believe that $f$ is not continuous at $z=0$, which would imply that $f$ is not differentiable at $z=0$. However, I'm not sure how I can prove this. 
Any help would be greatly appreciated. Thank you!!
EDIT:
I now see that $f$ is continuous at $z=0$, but I'm not sure how to prove that it is not differentiable.  I know I should use $\lim_{z\to 0} \frac{f(z) - f(0)}{z-0}$ to determine differentiability, and that I should get two different answers when I approach the limit from two different paths (for example: $x=0$, $y=0$, or $x=y$), but I'm not sure how to evaluate $\lim_{z\to 0} \frac{f(z) - f(0)}{z-0}$ for $x=0$, $y=0$ or $x=y$.
As before, any help would be much appreciated. Thanks! 

HINT: Are the limits as $x \to 0$ and $y \to 0$ the same? Because if the limit does not exist and equal the same value for EVERY direction of approach to the origin, then the limit does not exist there.
EDIT: The same argument would work with directional derivatives at the origin; if any two are not equal, then the function (considered as a function on R^2), can not be differentiable and hence whether or not the Cauchy-Riemann equations hold is insufficient.
However, do note that even if all directional derivatives exist and are equal, this is still not sufficient to prove the differentiability at the origin; it is only a necessary, but not a sufficient condition.
