How to find functional square root of $\sin(x)$.

Sebapi's idea is fine, but he has miscalculations. The coefficients of $f$ can be computed recursively with elementary finite computations. We start with $$\sin x=x-{1\over3!}x^3+{1\over5!}x^5-{1\over7!} x^7+{1\over9!}x^9+{1\over11!}x^{11}\ldots\tag{1}$$and make the Ansatz $$f(x):=\sum_{k=1}^\infty a_k x^k\ .$$ Putting the subsequent coefficients of $f\bigl(f(x)\bigr)-\sin x$ to zero we first have to solve $a_1^2=1$. I choose $a_1=1$ (and leave $a_1=-1$ to you). It then becomes quickly apparent that all $a_{2j}=0$, so that it seems reasonable to replace the above Ansatz by $$f(x):=\sum_{j=1}^\infty a_{2j-1} x^{2j-1}\ .$$ I did the computations with Mathematica, and obtained $$f(x)=x-{1\over12}x^3-{1\over160}x^5-{53\over40\,320}x^7-{23\over71\,680}x^9-{92\,713\over1\,277\,337\,600}x^{11}+\ \ldots\ .\tag{2}$$ Computing $f\bigl(f(x)\bigr)$ up to the $x^{11}$ term using $(2)$ gives exactly $(1)$. The numerators appearing in $(2)$ are listed at OEIS under A098932, where reference is made to the problem at hand.