Why is $y = x^{1/3}$ not differentiable at $x = 0$? It seems like the differentiability of a curve should morally speaking be part of the curve itself. More specifically, whether or not a curve is differentiable should not depend on the coordinates.
So the problem is that $x^{3}$ is differentiable everywhere and $x^{1/3}$ is not, even though the latter is just the former in different coordinates. Why should a function lose differentiability just by changing coordinates?
 A: There are two different points in your question that I think need to be answered:


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*Is $y'(x)=\frac{y'(t)}{x'(t)}$ invariant for a change of variable?
No, it isn't: the slope of the tangent vector is dipendent on your frame of references. Even simply rotating it can change your result, leading to the non existence of the ratio (that is, to a vertical vector, which is not even that patological). Changing the coordinates by something that is not a diffeomorphism changes the results too (for more information, look at the notes)
You adress as regularity this ratio, but it is not; A curve can be regular even if its tangent vector is vertical. As an example, think of the line $y=3$: it is indeed quite regular, even thought it cannot be written as $y=mx+q$

Note 1
You might now be curious if the effective regularity can be "damaged" by a change of coordinates: I will try to answer to this question


*Does a simple change of coordinates modify the regularity of the curve itself?
The question is not so easy to answer: if you allow any kind of parametrization then yes, it does, as you example shows (changing $t'=t^{\frac{1}{3}}$, or making the inversion $y'=x,\ x'=y$). On the other hand, curves are usually studied with a special parameter, called arc-length or natural parameter (see here for more informations) that avoids this problem: equipped with such a parameter, the tangent vector has always length $1$ (if the curve satisfies certain requirements, such as being $C^1$, $\gamma'\neq 0$, not self intersecting and so on), and the regularity of the curve is not changed by a diffeomorfism
As for your example, the regularity is indeed different: if $y=|x|$ gets parametrized by arc lenght, we get: $\gamma(t)=\frac{1}{\sqrt{2}}(t,|t|)$, and this curve does not have a tangent vector defined in $t=0$, so it is indeed non regular.
Computing the vector tangent to $y=x^{\frac{1}{3}}$ with the natural parameter, we get $\dot{\gamma}(0)=(0,1)$
Note 2 (on the natural parameter):
As user647486 pointed out, in more than $1$ dimension a set can be equipped with a lot of different differentiable structures (see here for an example by John Milnor). In one dimension, however, the differentiable (and metric) structure is unique: this is (for the metric structure) due to the fact that we are able to make a regular curve isomorphic to a interval, thing that we are un-able to do in more than one dimension, as the Theorema Egregium shows
