How to prove a functional equation has finite dimensional solution space? Consider a functional equation for meromorphic functions $S(z)$, such as the following:
$$S(z^2) + w S(wz^2) + w^2 S(w^2z^2) = 3zS(z^3)-3zS(z^6),$$
where $w=e^{i2\pi/3}$. It is obvious that all solutions form a linear space which contains $S(z)=1$ and $S(z)=\tfrac{1}{1-z}$. Are there any tools to prove or disaprove this linear space is finite dimensional?
 A: Tl;Dr - The OP appears to have found the two function that span the space, except the result appears to rely on a variation of the Collatz Conjecture. If any number theorists can help clarify it would be appreciated.

We start by stipulating we get a series about the origin
$$S(z) = \sum_{\mathbb Z} a_kz^k$$
Plugging this into the relation, we get
$$\sum_{\mathbb Z} a_kz^{2k}+w\sum_{\mathbb Z} a_k w^k z^{2k} + w^2\sum_{\mathbb Z} a_k w^{2k} z^{2k}=3z\sum_{\mathbb Z} a_kz^{3k}-3z\sum_{\mathbb Z} a_kz^{6k}$$
$$\sum_{\mathbb Z} a_kz^{2k}(1+w^{k+1}+w^{2k+2})=3z\sum_{\mathbb Z} a_k(z^{3k}-z^{6k})$$
Note $1+w^{k+1}+w^{2k+2}$ is $3$ if $k = 3n+2$ for some $n \in \mathbb Z$, else is $0$.
$$3\sum_{\mathbb Z} a_{3k+2}z^{2(3k+2)} = 3z\sum_{\mathbb Z} a_k(z^{3k}-z^{6k})$$
basic algebra and letting $z^3 \mapsto z$ yields
$$\sum_{\mathbb Z} a_{3k+2}z^{2k+1} = \sum_{\mathbb Z} a_k (z^k-z^{2k})$$
Expanding this, we find
$$a_2 z + a_5 z^3 + a_8 z^5 + \cdots +\\= a_1 (z - z^2) + a_2 (z^2 - z^4) + a_3 (z^3 - z^6) + a_4 (z^4 - z^8) + a_5 (z^5 - z^{10}) + a_6 (z^6 - z^{12}) +\cdots +$$
Rearranging, we have
$$a_2 z + a_5 z^3 + a_8 z^5 + \cdots +\\= a_1 z + (a_2 - a_1) z^2 + a_3 z^3 + (a_4 - a_2) z^4  + a_5 z^5 + (a_6 - a_3) z^6 + \cdots +$$
We thus get the two relations $a_{2k+1} = a_{3k+2}$ and $a_{2k} = a_k$. My classical number theory is very weak, but numerical testing and some quick heuristics about prime factorizations show this implies all coefficients but the first must be equal. Seeking $S(z)$ holomorphic at the origin, we thus have
$$\bbox[8px,border:1px solid black]
{S(z) = a_0 + \sum_{k=0}^\infty a_1 z^k = a_0 + \tfrac{a_1}{1-z} = \operatorname{Span}\left(1,\tfrac{1}{1-z}\right)}\;\;$$
It is quick to check that starting the series from any negative powers, regardless of coefficient, does not satisfy the relation so we begin at $k=0$
All that remains is to prove the number theory conjecture, which appears related to the Collatz Conjecture 
(which has $a_{2n} = a_n$ and $a_{2n+1} = \color{red}{a_{6n+4}}$) and may thus be untractable at the moment.
