How are the sheets of $y^3 = x(x-1)(x+1)$ arranged? When I have a hyperelliptic curve, such as $X = \{ (x,y): y^3 = x(x^2 -1)\} \subseteq \mathbb{C}^2 $.  This is a 3-fold branched cover of the sphere at the double-points, $x \in \{  0, 1, -1 \}$ using the map $p:(x,y) \mapsto x$.
So there is a map $\pi_1( \mathbb{C} \backslash \{ 0, 1, -1\}) \to S^3$ generated by 3 circles, describing $X$ e.g. 
$$ C =  \big\{ x_0 + r \, e^{i\theta } : \theta \in [0, 2\pi] \big\} \in \pi_1(X) $$
I'd like to know how the three sheets are arranged in this case. How are the pre-images permuted? $(x, \omega \sqrt[3]{x\,(x +1)(x-1)})$ with $\omega^3 = 1$.

Notes
Any curve of the form $y^d = (x-b_1) \dots (x - b_n)$ defines a Riemann surface with branch points at $b_1, b_2, \dots, b_n$.
The Riemann surface $y^3 = (x-a)(x-b)(x+c)$ is related to integrals such as:
$$ I(a,b,c) = \int_0^1 \frac{dx}{\sqrt[3]{(x-a)(x-b)(x+c)}} $$
 A: I'm not quite sure what your $S^3$ is. Also, since you consider this a covering of the Riemann sphere, your curve really lives in $\mathbb{P}^2(\mathbb{C})$. In homogeneous coordinates it's 
$$
\{[x:y:x]: y^3 = x(x-z)(x+z)\}
$$
and the branched covering map $X \rightarrow \mathbb{P}^1$ is $p: [x:y:z] \mapsto [x:z]$. It is unbranched over $[1:0]$ which is "the point at infinity" of the Riemann sphere downstairs. (Exercise: for which $(d, n)$ is $y^d = (x-b_1)\dots(x-b_n)$ unbranched at infinity?) Having determined that the only branch points are in the affine patch $z\neq 0$ we may revert to affine coordinates and put $z=1$. 
Now, we are interested in the variation of the argument of the complex number $x(x-1)(x+1)$ as we walk around each branch point. The argument is itself not single-valued but for any choice of branch it remains true that
$$
\operatorname{Arg}\prod_i z_i \equiv \sum_i \operatorname{Arg} z_i \pmod{2\pi},
$$
which is good enough for our purpose. Walking around $x=1$ (say) in a small circle, you find that 
$$
\operatorname{Arg} \left(x(x-1)(x+1)\right) \equiv \operatorname{Arg} x + \operatorname{Arg} (x-1) +\operatorname{Arg} (x+1)  \pmod{2\pi},
$$
and the terms from $x$ and $x+1$ remain approximately constant, while $\operatorname{Arg} (x-1)$ increases continuously by $2\pi$. (It will drop back since $\operatorname{Arg}$ has been made single-valued but it's the continuous variation we're after.) Thus, if you attempt to extract a continuous cube root, you'll pick up an argument of $+2\pi/3$ going around the branch point counterclockwise. Moreover, you'll see that this really only uses the fact that the branch point was a simple root of the right-hand-side of your equation. Thus we have actually dealt with all three of them.
Next we need to choose branch cuts and label the sheets. Every branch point must connect to a cut because we are trying to make $y$ single-valued away from the cuts, and we have seen that going around a branch point will multiply $y$ by $e^{2\pi i/3}$. Therefore every circle around a branch point must meet at least one cut. Apart from that, we have a great deal of freedom. I will cut along the real axis from $x=-1$ to $x=1$ (so that we have two cuts coming out from $x=0$, which is OK).
Let me call the sheet that has $y = re^{2\pi i k/3}$ for real $x>1$ "sheet number $k$", where $k = 0, 1, 2$. (Not feeling imaginative today.)
We saw already that going counterclockwise around any one branch point induces the permutation $0\rightarrow 1\rightarrow 2\rightarrow 0$. Now this information gets encoded as identification of sheets across the branch cuts. So, going down across the cut between $x=0$ and $x=1$ is just the permutation already mentioned. Going around $x=-1$ counterclockwise amounts to going up across the cut from $x=-1$ to $x=0$. Thus, going down across that cut induces the inverse, $0\rightarrow 2\rightarrow 1\rightarrow 0$. This determines what happens as you go around $x=0$, in which case you cross both cuts. 
By the way, curves admitting a $3:1$ map to $\mathbb{P}^1$ are called trigonal, this being where the terminology stabilizes: birational, hyperelliptic, trigonal, tetragonal...
A: Around $x = 0$, $f(x, y) = y^3 - x(x^2 - 1)$ is asymptotically $y^3 + x$, and the Puiseux expansion will be in the powers of $(-x)^{1/3}$. In terms of the Newton polygon, there are only two vertices closest to the origin that contribute to the expansion.
The structure at $x = -1$ and at $x = 1$ is the same: $f(\pm 1 + x, y) \sim y^3 - 2x$, and the expansion of $f(x, y)$ will be in terms of $(\pm 1 + x)^{1/3}$.
Around infinity, we have $f(x, y) \sim y^3 - x^3$; the expansion will be in terms of $x$, and $f(x, y)$ is unbranched over $\infty$.
Thus all permutations are the cycles $(1 2 3)$, and the group is $\mathrm C_3$. The structure of the branches looks like this:

