Frechet differential in $l^\infty$ 
Let $(f_n)_{n \in \mathbb{N}}$ with $f_n \in \mathcal{C}^1(\mathbb{R})$ for all $n$ and assume that that $(f_n)_n$ and $(f^\prime_n)_n$ are both uniformly bounded and equicontinuous on every $[a,b] \subset \mathbb{R}$.
Show that the operator $F: l^\infty \rightarrow l^\infty$ with $F((x_1,x_2,\ldots)) = (f_1(x_1),f_2(x_2),\ldots)$ is well-defined and Frechet differentiable and find the Frechet differential $DF$.

Here is my attempt: Since the $f_n$ are uniformly bounded we have that $F((x_1,x_2,\ldots)) =(f(x_1),f(x_2),\ldots)$  is bounded, so $F$ indeed maps $l^\infty$ to $l^\infty$ and is thus well-defined.
Furthermore, if I am not mistaken, since the $f_n$ are working over $\mathbb{R}$ the Frechet derivative is identical to the common derivative. So
\begin{align}
DF(x) &= \lim_{t \rightarrow 0} \frac{F(x+tv)-F(x)}{t} \\ &= \lim_{t \rightarrow 0} \frac{(f_1(x_1+tv_1),f_2(x_2+tv_2),\ldots)-(f_1(x_1),f_2(x_2),\ldots)}{t}\\ &= (f_1^\prime(x_1),f_2^\prime(x_2),\ldots)
\end{align}
, where we used the condition of equicontinuity in the last equation. Is this correct?
 A: Here is a solution using directional derivatives as the OP had tried.
I will  denote elements in $\ell_\infty$ as functions $\mathbf{x}:\mathbb{N}\rightarrow\mathbb{R}$ with $\|\mathbf{x}\|_\infty:=\sup_n|\mathbf{x}(n)|<\infty$
By hypothesis, on any bounded interval$(a,b)$, $M:=\sup_n(\|f_n\|_u+\|f'_n\|_u)<\infty$;
and for any $\varepsilon>0$, there is $\delta>0$ such that $\sup_n\big(|f_n(u)-f_n(v)|+|f'_n(u)-f'_n(v)|\big)<\varepsilon$ whenever $u,v\in(a, b)$ and $|u-v|<\delta$. (Here $\|\;\|_u$  is the uniform in the corresponding interval $(a, b)$).
Fix $\mathbf{x}_0$ and consider the ball $B_0=B(\mathbf{x}_0;1)$. Any $\mathbf{x}\in B_0$ satisfies $\mathbf x(n)\in (-1-\|\mathbf{x}_0\|_\infty,1+\|\mathbf{x}_0\|_\infty)=(a, b)$
It follows that $F:\mathbf{x}\mapsto\Big(f_n\big(\mathbf{x}(n)\big):n\in\mathbf{N}\Big)$ is a well defined map from $\ell_\infty$ into $\ell_\infty$.
For each $\mathbf{x}\in B_0$ define the linear map
$$A_{\mathbf{x}}:\mathbf{h}\mapsto\big(f'_n(\mathbf{x}(n))\,\mathbf{h}(n):n\in\mathbb{N}\big)$$
Notice that
$$|\big(A_{\mathbf{x}}\mathbf{h}\big)(n)|\leq M\|\mathbf{h}\|_\infty,\qquad n\in\mathbb{N}.$$
Also, for any $\mathbf{x},\mathbf{y}\in B(\mathbf{x}_0;1)$ with $\|\mathbf{x}-\mathbf{y}\|_\infty<\delta$,
\begin{align}
|\big((A_\mathbf{x}-A_{\mathbf{y}})\mathbf{h}\big)(n)|&\leq |\big(f'_n((\mathbf{x}(n))-f'_n(\mathbf{y}(x)\big)\mathbf{h}(n)|\\
&\leq\big|f'_n((\mathbf{x}(n))-f'_n(\mathbf{y}(n))\big|\|\mathbf{h}\|_\infty<\|\mathbf{h}\|_\infty\varepsilon
\end{align}
by uniform equicontinuity assumption. Hence $\mathbf{x}\rightarrow A_\mathbf{x}$ is a continuous map from $B_0$ into $L(\ell_\infty,\ell_\infty)$.
For $\mathbf{x}\in B_0$ and $\mathbf{h}\in\ell_\infty$,  if $|t|\|\mathbf{h}\|_\infty<\delta$
\begin{align}
\Big|\frac{\big(F(\mathbf{x}+t\mathbf{h})-F(\mathbf{x})\big)(n)-t\big(A_{\mathbf{x}}\mathbf{h}\big)(n)}{t}\Big|&=\left|\Big(f'_n\big(\mathbf{x}(n)+\theta_nt\mathbf{h}(n)\big)-f'_n\big(\mathbf{x}(n)\big)\Big)\mathbf{h}(n)\right|\\
&\leq \Big|f'_n\big(\mathbf{x}(n)+\theta_nt\mathbf{h}(n)\big)-f'_n\big(\mathbf{x}(n)\big)\Big|\|\mathbf{h}\|_\infty<\varepsilon\|\mathbf{h}\|_\infty
\end{align}
($\theta_n\in(0,1)$ for all $n\in\mathbb{N}$) by the equicontinuity of $f'_n$ and the mean value theorem. Hence
$$\lim_n\left\|\frac{F(\mathbf{x}+t\mathbf{h})-F(\mathbf{x})}{t}-A_\mathbf{x}\mathbf{h}\right\|_\infty=0$$
We have shown that $F$ is Gateâux differentiable at any point $\mathbf{x}\in\ell_\infty$, and that the Gateâux derivative is continuous. Therefore, $F$ is Frechét differentiable, and
$F'(\mathbf{x})\mathbf{h}=A_\mathbf{x}\mathbf{h}$ for all $\mathbf{x},\mathbf{h}\in\ell_\infty$.

Edit: A direct approach is as follows. With the notation set out above,  the uniform equicontinuity assumption implies that for  $\mathbf{h}$ with $\|\mathbf{h}\|<\min(\delta,1)$
\begin{align}
\Big|\big(F(\mathbf{x}_0+\mathbf{h}\big)(n)-\big(F(\mathbf{x}_0)\big)(n)-\big(A_{\mathbf{x}_0}\mathbf{h}\big)(n)\Big|&=
\Big|f_n\big(\mathbf{x}_0(n)+\mathbf{h}(n)\big)-f(\mathbf{x}_0(n))-f'_n(\mathbf{x}_0(n))\mathbf{h}(n)\Big|\\
&=\left|\Big(f'_n\big(\mathbf{x}_0(n)+\theta_n\mathbf{h}(n)\big)-f'_n(\mathbf{x}_0(n))\big)\mathbf{h}(n)\right|\\
&<\varepsilon\|\mathbf{h}\|_\infty
\end{align}
where the $0<\theta_n<1$  (application of the mean value theorem). Hence
$$\lim_{\mathbf{h}\rightarrow\boldsymbol{0}}\frac{1}{\|\mathbf{h}\|_\infty}\big\|F(\mathbf{x}_0+\mathbf{h})-F(\mathbf{x}_0) -A_{\mathbf{x}_0}\mathbf{h}\big\|_\infty=0$$
that is,   $F$ is Fréchet differentiable at $\mathbf{x}_0$ and that $F'(\mathbf{x}_0)=A_{\mathbf{x}_0}$.
