A rational rabbit

Choose $c$ to the the unique complex number with positive imaginary part such that $f_c(z)=z^2+c$ has a super-attractive orbit of period 3. Suppose we conjugate $f_c$ to obtain a rational function $g$ with the property that
$$0\to1\to\infty\to0$$
forms a super attractive orbit. Generate the Julia set of $g$.

I'll answer the first part.

Define the functions

f[c_][z_] := z^2 + c

and

F[c_][z_] := Nest[f[c], z, 3]

To find the critical points of $F_c(z)$, solve $F_c^\prime(z)=0$ for $z$. I used the code

z /. Solve[D[F[c][z], z] == 0, z]

To determine a list of $c$ values such that each critical point $z$ is also a fixed point, solve $F_c(z)=z$ for $c$. After deleting duplicate $c$ values and rejecting those $c$ values for which the imaginary part is not positive, we are left with only one possible value of $c$, $c=-0.122561+0.744862i$. This entire process is achieved by the code

First[
 Select[
  DeleteDuplicates[
   Flatten[
    c /. NSolve[F[c][#] == #, c] & /@ %
   ],
   Abs[#1 - #2] < 10^-10 &
  ],
  Im[#] > 0 &
 ]
]

More simply, since a super attractive orbit must contain a critical point, we can solve $f_c^\prime(z)=0$ for $z$ using the Mathematica code

z /. Solve[D[f[c][z], z] == 0, z]

and proceed as before.

@RedCrayon That's it alright. With that in hand, you should be able to identify the points in the orbit, say
$$z_0 \to z_1 \to z_2 \to z_0.$$
That's because we know we can take $z_0=0$ and then use $f$ to compute $z_1$ and $z_2$.

Once we know $z_0$, $z_1$, and $z_2$, we should be able to find a Mobius transformation sending
$$z_0\to 0, z_1 \to 1, \text{ and }\, z_3 \to \infty.$$

If we conjugate $f_c$ with this Mobius transformation, we should get the function we are after.

And don't forget that we have a Mathematica notebook to help us find Mobius transformations to send 3 points to 3 points here.

We can compute $z_0$, $z_1$, and $z_2$ with the NestList function: these are $z_0=0$, $z_1=-0.1225611669 + 0.7448617666 i$, and $z_2=-0.6623589786 + 0.5622795121 i$.

A mobius transformation sending $z_0\rightarrow0$, $z_1\rightarrow1$, and $z_2\rightarrow\infty$ is $$\phi(z)=\frac{(z_1-z_2)z}{z_1z-z_1z_2}.$$

The inverse is $$\phi^{-1}(z)\frac{z_1z_2z}{z_1z-z_1+z_2}.$$

In Mathematica, if we define $\phi$

z1 = -0.1225611669 + 0.7448617666*I;
z2=-0.6623589786 + 0.5622795121*I;
Phi[z_] := ((z1 - z2) z)/(z1 z - z1 z2);

then $\phi^{-1}$ is

InvPhi[z_] = First[w /. NSolve[Phi[w] == z, w]]

We define $g=\phi\circ f\circ\phi^{-1}$, or in Mathematica,

c0 = -0.122561+0.744862*I;
g[z_] = Phi[f[c0][InvPhi[z]]]

The resulting function is complicated and difficult to describe succinctly, but we can generate the Julia set...