Consider Banach spaces $X, Y_i, Z$ and let $U \subseteq X$, $V \subseteq Y_1 \times \ldots \times Y_n$ and let $f:= (f_1, \ldots, f_n): U \rightarrow V$ and $g: V \rightarrow Z$.
Assume that $f$ is differentiable in $x \in U$ and $g$ is differentiable in $f(x)$. Compute the Frechet derivative $D(g \circ f)_x(h)$ in terms of $f_1, \ldots, f_n, g$.
I realise that we need the chain rule here, so
$$D(g \circ f)_x = Dg_{f_x} \circ Df_x$$
However, I do not see how to continue from here. If the whole thing was over $\mathbb{R}^n$ I recognise that we would have Jacobi matrices, but since the setting is over Banach spaces I am unsure what to do here.
Not sure if this is what are you looking for, but:
If $\pi_i \colon Y_1 \times \cdots \times Y_n \to Y_i$ is the projection onto the $i$-th coordinate, then $f_i = \pi_i \circ f$, so $$ D(f_i)_x = D(\pi_i)_{f(x)} \circ Df_x = \pi_i \circ Df_x. $$ (If $T \colon E \to F$ is a linear map between Banach spaces, then $DT_v=T$ for every $v \in E$.) Hence $$ Df_x = (D(f_1)_x,\dots,D(f_n)_x) $$ and this allows to write $D(g \circ f)_x$ in terms of (the Fréchet-derivatives of) $f_1,\dots,f_n$ and $g$.