As is typically done in iterated systems, let $f(0)=0$ where $f(z)$ is smooth. Then applying Riordan's approach to solving inverse Bell polynomials produces the derivatives of $f(z)$ which are used for the power series $f(z)$ giving $$f(z)=\sum_{k=1}^{\infty}\frac{1}{k!}f^{(k)}(0) z^k.$$ There are two solutions based on whether $f'(0) = -\sqrt{g'(0)}$ or $f'(0) = \sqrt{g'(0)}$. Due to space, only the first three derivitives are shown.
Mathematica Code
f[0]=0;
max = 3;
Solve[Table[D[g[z] == f[f[z]], {z, i}] /. z -> 0 , {i, max}],
Table[D[f[z], {z, i}] /. z -> 0, {i, max}]]
Solution 1 $$f'(0) = -\sqrt{g'(0)}$$ $$f''(0) = \frac{g'(0) g''(0)+\sqrt{g'(0)} g''(0)}{g'(0)^2-g'(0)} $$ $$f'''(0) = -\frac{3 g''(0)^2+g^{(3)}(0) \sqrt{g'(0)} \left(\sqrt{g'(0)}-1\right)^2}{\left(\sqrt{g'(0)}-1\right)^2 g'(0) \left(g'(0)+1\right)}$$
Solution 2
$$f'(0) = \sqrt{g'(0)} $$ $$f''(0) = \frac{g'(0) g''(0)-\sqrt{g'(0)} g''(0)}{g'(0)^2-g'(0)} $$ $$f'''(0) = \frac{g^{(3)}(0) \left(\sqrt{g'(0)}+1\right)^2 \sqrt{g'(0)}-3 g''(0)^2}{\left(\sqrt{g'(0)}+1\right)^2 g'(0) \left(g'(0)+1\right)}$$
Riordan, John, Combinatorial identities, New York- London-Sydney: John Wiley and Sons, Inc. 1968. XII, 256 p. (1968). ZBL0194.00502.