For $q \in \Bbb C$, let $f_q(x) = \frac {3x + q(1-x)}{1 - q(1-x)}$
and for $n \in \Bbb N$, let $F_n(q) = (f_1 \circ f_q \circ f_{q^2} \circ \cdots \circ f_{q^n}) (0)$
For $|q| \notin \{ 1, \frac 13 \}$, the sequence $(F_n(q))$ converges to some $F(q) \in \Bbb P^1(\Bbb C)$, and it looks like $F$ is meromorphic there (should not be too difficult to prove)
Here is a picture of $F_{100}$ :
(surprisingly, it looks meromorphic on most of the unit circle, save some essential singularities, but it's almost surely deceiving. There should be a necklace of zeros and poles around the circle, but you don't see them because they are so close together)
I am more interested in what happens for $|q| < \frac 13$, where $f_{q^n}$ converges too fast to the homothety $x \mapsto 3x$ for the argument $0$ to escape very far. Looking at the picture, it looks very continuable.
Thankfully, I picked some coefficients so that the Laurent series of $F_n$ at $0$ converge in $\Bbb Z((q))$, which gives a first way to see past the $|q| = \frac 13$ boundary.
Using the series $F(q) = \frac 1q(1-2q+4q^2-6q^3+20q^4-46q^5+\cdots)$ and truncating after a hundred terms, I get another picture
The radius of convergence here seems to be around $0.41$, and it doesn't look any less well-behaved except for possibly a pole at $q = -0.41$.
Pre and post composing the function with carefully chosen functions can also extend this even further.
So the "natural" boundary of $|q|= \frac 13 $ isn't really one, and still it looks like it could be extended further, possibly up to the whole open disk of radius $1$.
So, what gives ? Is there an alternative way of computing the extension in the annulus $\frac 13 \le |q| < 0.41$ that would explain it and gives a further continuation ?