How is the function$ f\left ( z \right )=\left ( x^{3}+3xy^{2}-3x \right )+i\left ( y^{3}+3x^{2}y-3y \right )$ nowhere analytic In a problem I've been doing it is stated to show that even though the function is differentiable along the coordinate axes, it is nowhere analytic. By definition, a function is differentiable if the following limit exists and is finite
$$\lim _{\Delta z \to0 }\frac{f\left ( z+\Delta z \right )-f\left ( z \right )}{\Delta z}.$$ 
Along the x-axis I got something like $f^{'}\left ( z \right )=3x^{2}+3x$(it's important that it exists). Similar analysis holds for the y-axis. The Cauchy-Riemann equations yield the following system :
$$x^{2}+y^{2}=y^{2}+x^{2}$$
$$xy=-xy$$
That means the the necessary conditions are met only along the lines $x=0$ or $y=0$. The partials of $u$ and $v$ are elementary functions and as such are continuous on their whole domain(which includes the axes).  I have, thus proven that the function IS differentiable along the axes, which is contrary to the problem's statement. This seemed a bit odd, so I tried doing this analysis by definition, namely, after proving differentiability(the difference between analyticity and differentiabilty is subtle) I used the following two definitions:

Definition 1. A function is analytic in a point if it is differentiable in SOME neighborhood of that point.

This obviously holds true for every point on the axes. 

Definition 2.  A function is analytic in some region if it is analytic in every point of that region. 

This intuitively also holds true. One point where this might break down is the fact that the axes might not even be considered regions(which are open here, in the topological sense.). By all definitions I have read and searched for, a line is to be considered an open region. Please, feel free to prove me wrong, as I have been thinking this over for a long period.

In order to be analytic at a point $z_0$, it must be complex-differentiable at every point in some open disk centered at $z_0$. (This is your definition $1$).
You've shown using the Cauchy-Riemann equations that this open disk condition doesn't hold at any $z_0$, including $z_0$ on the coordinate axes. Hence $f(z)$ is analytic nowhere.
