an interesting q-series and a certain continued-fraction My aim is to find a rigorous proof of the following conjectured identity.Given 
$$1+q+q^2-q^4-q^5+q^7+q^8-q^{10}-q^{11}+\ddots=\cfrac{1}{1-q+\cfrac{(q^3)}{1-q^3+\cfrac{q^2(1+q)(1+q^3)}{1-q^5+\cfrac{q^3(1+q^2)(1+q^4)}{1-q^7+\cfrac{q^4(1+q^3)(1+q^5)}{1-q^9+\ddots}}}}}$$
The exponents on the left-hand side $${1,2,4,5,7,8,\ddots}$$ are positive integers not divisible by $3$.
How do we show that, the identity is true?
Independent verification and attempts at proving/disproving the result are very welcome. 
 A: Regarding R. Israel's remark, three of your continued fractions, while not exactly the same, are variants of a common form discussed here for $|q|<1$,
$$\frac{1}{1-q} =\cfrac{1}{1+q-\cfrac{\color{brown}{2q(1+q^2)}}{1+q^3+\cfrac{q^2(1-q)(1-q^3)}{1+q^5-\cfrac{q^3(1+q^2)(1+q^4)}{1+q^7+\cfrac{q^4(1-q^3)(1-q^5)}{1+q^9-\ddots}}}}}\tag0$$
First, the one for the Fibonacci numbers in this post,
$$\chi(q)=\cfrac{1}{1+q-\cfrac{\color{brown}{(1+q^2)}}{1+q^3+\cfrac{q^2(1-q)(1-q^3)}{1+q^5-\cfrac{q^3(1+q^2)(1+q^4)}{1+q^7+\cfrac{q^4(1-q^3)(1-q^5)}{1+q^9-\ddots}}}}}\tag1$$
Second, the one for the Narayana numbers in this post,
$$\psi\Big(q\Big)=\cfrac{1}{1+q-\cfrac{\color{brown}{q^2}}{1+q^3+\cfrac{q^2(1-q)(1-q^3)}{1+q^5-\cfrac{q^3(1+q^2)(1+q^4)}{1+q^7+\cfrac{q^4(1-q^3)(1-q^5)}{1+q^9-\ddots}}}}}\tag2$$
Third, the one for this post with $q \to -q$,
$$M(q) = \cfrac{1}{1-q+\cfrac{\color{brown}{q^3}}{1-q^3+\cfrac{q^2(1+q)(1+q^3)}{1-q^5+\cfrac{q^3(1+q^2)(1+q^4)}{1-q^7+\cfrac{q^4(1+q^3)(1+q^5)}{1-q^9+\ddots}}}}}\tag3$$
Arranged in this way, one can see their common form. Notice that the brown part is the only "level" that changes, as well as letting $q\to-q$. (Based on this observation, I experimented with using $\color{brown}{-1+q^2, -q+q^2, q+q^2, q^4}$ instead of the ones above, and they also yield generating functions of known sequences in the OEIS.)
P.S. Transforming $(0)$ to $(3)$, I get,
$$\begin{aligned}M(q)
&=\frac{1+q^2}{1-q+q^2}\\ 
&= 1+q+q^2-q^4-q^5+q^7+q^8-q^{10}-q^{11}+\dots
\end{aligned}$$ 
and the sequence is A163806 (the Gf is corrected by B. Berselli). It states that,
$$a(3n) = 0$$
thus your observation that the exponents with non-zero coefficients are the integers not divisible by 3 is correct. (Note that this formula is only valid for $|q|<1$.)
