I find the strong Fréchet derivative of $\Phi(h,\psi(h))$, where $\Phi:T_0\times T_\xi\to Y$ with $T_0, T_\xi, Y$ Banach spaces and $\psi:T_0\to T_\xi$ is strongly differentiable in $0$, evaluated in $h=0$, as $\Phi'_h(0,0)+\Phi_{\xi}'(0,0)\psi'(0)$ in Kolmogorov-Fomin's Элементы теории функций и функционального анализа, p. 499.
I think there is some chain rule similarly to the case of $F:\mathbb{R}^2\to\mathbb{R}$ differentiable in $(x,y)$ and $f,g:\mathbb{R}\to\mathbb{R}$ differentiable in $t$, for which we have the equality $$\frac{d}{dt}F(f(t),g(t))=f'(t)\frac{\partial }{\partial x}F(f(t),g(t))+g'(t)\frac{\partial }{\partial y}F(f(t),g(t))$$
Nevertheless, I know a proof of this latest equality using matrices, which I cannot adapt to verify whether something analogous holds for strong derivatives of maps between Banach spaces. How can an equality such as $\frac{d}{dh}\Phi(h,\psi(h))|_{h=0}=\Phi'_h(0,0)+\Phi_{\xi}'(0,0)\psi'(0)$ be justified? Does soemthing like the chain rule I quoted applies? If it does, how can it be proved? I only know that if $X,Y,Z$ are Banach spaces, $F:X\to Y$ is strongly differentiable in $x_0$ and $G:Y\to Z$ is strongly differentiable in $y_0=F(x_0)$, then $(G\circ F)'(x_0)=G(F(x_0))F(x_0)$. Thank you very much!!!