Too long for a comment. I find your question interesting for a reason linked to the proof of the Faà de Bruno formula. Let $f,g$ be two functions from $\mathbb R$ into itself. Let me assume that $f(0)=0$ and let me give a formal expression for $(g\circ f)(x)$ using the Fourier transform of $g$:
$$
g(f(x))=\int \hat g(\eta) e^{2iπ \eta f(x)} d\eta=\sum_{k\ge 0}\int \hat g(\eta)\frac{(2iπ \eta f(x))^k}{k!}d\eta,
$$
so that
$$
g(f(x))=\sum_{k\ge 0}\int \hat g(\eta)\frac{(2iπ \eta)^k}{k!}\prod_{1\le j\le k}\hat f(\xi_j) e^{2iπ x\xi_j}d\xi_j
d\eta
\\=
\sum_{k\ge 0}\int \hat g(\eta)\frac{(2iπ \eta)^k}{k!}\prod_{1\le j\le k}\hat f(\xi_j) \sum_{m\ge 0}\frac{(x2iπ \sum \xi_j)^m}{m!}
d\xi
d\eta,
$$
and after this iteration of Fourier transformations this yields for $m\ge 1$ the Faà de Bruno Formula,
$$
\frac{(g\circ f)^{(m)}(0)}{m!}=\sum_{k\ge 1\atop \sum_{1\le j\le k} \alpha_j=m}
\frac{g^{(k)}(f(0))}{k!}\prod_{1\le j\le k}\frac{f^{(\alpha_j)}(0)}{\alpha_j!}.
$$