Prove that $\ddot {g} (t) = D^2 f(x+th) (h,h).$ 
Let $E$ be a Banach space. Let $U \subseteq E$ be open and $f : U \longrightarrow F$ be $C^2.$ Let $x \in U,h \in E$ be such that $B(x, \|h\|) \subseteq U.$ Consider the map $g : (-1,1) \longrightarrow F$ defined by $g(t) = f(x+th),\ t \in (-1,1).$ Then $\ddot {g} (t) = D^2 f(x+t h) (h,h).$

I have proved that $\dot {g} (t) = Df (x + t h) (h).$ In order to find the $\ddot {g} (t)$ I need to compute the limit $$\lim\limits_{s \to 0} \frac {Df (x + th + sh) (h) - Df (x + th) (h)} {s}.$$
How to simplify the above expression? Since $Df$ is Frechet differentiable I know that $$\left \|Df(x + k) - Df (x) - D^2 f(x) (k) \right \| = o(\|h\|).$$
I adopted the following strategy $:$
I know that the above limit exists. Let it be $y \in F.$ But then $$\lim\limits_{s \to 0} \frac {\left \|Df(x + th + sh)(h) - Df(x+th) (h) \right \|} {|s|} = \|y\|.$$ Now $$\begin{align*} \left \|Df(x +th +sh)(h) - Df(x+th)(h) \right \| & = \left \|(Df(x + th + sh) - Df(x + th)) (h) \right \| \\ & = \left \|\left (Df(x + th +sh) - Df(x + th) - D^2f (x+th) (sh) \right )(h) + D^2 f(x + th) (sh) (h) \right \| \\ & \leq \left \|Df(x + th + sh) - Df(x + th) - D^2 f(x + th)(sh) \right \| \|h\|  + \left \|s\ D^2f (x +th) (h)(h) \right \|  \\ & = o(\|sh\|)\|h\| + |s| \left \|D^2f(x + th) (h)(h) \right \| \\ & = o(|s|\|h\|) \|h\| + |s| \left \|D^2f(x + th) (h)(h) \right \| \end{align*}$$ This shows that $$\lim\limits_{s \to 0} \frac {\left \|Df (x + th + sh) (h) - Df (x + th) (h) \right \|} {|s|} \leq \left \|D^2f(x + th) (h) (h) \right \|.$$
How do I proceed now? Any help regarding this will be highly appreciated.
Thanks for your time.
 A: Your problem is you want to show that
\begin{equation*}
\lim_{s \to 0} \left\|\frac{Df(x + th + sh)(h) - Df(x + th)(h) - sD^{2}f(x + th)(h,h)}{s}\right\| = 0.
\end{equation*}
The fact that
\begin{equation*}
\lim_{s \to 0} \left\|\frac{Df(x + th + sh)(h) - Df(x + th)(h)}{s}\right\| = \|D^{2}f(x + th)(h,h)\|
\end{equation*}
is also true, but it's of no particular use here and it's getting you stuck.
This is an easy mistake to make when starting out with Fréchet derivatives.  The 1D habit to write $\frac{f(t + h) - f(t)}{h} \to f'(t)$ becomes a  bit clumsy and it becomes worthwhile to think of $\frac{f(t + h) - f(t) - f'(t)h}{h}$ going to zero instead.
Edit: At the request of the author, I'll show how to compute $\ddot{g}(t)$ using the chain rule.
$\dot{g}(t) = Df(x + th)(h)$.  This is a composition so the chain rule can be applied to find $\ddot{g}(t)$.  The issue is writing this coherently as a composition.
The first operation involved is scalar multiplication and vector addition $t \mapsto x + th$.  Next, application of $Df$: $y \mapsto Df(y)$.  Finally, evaluating the matrix so-obtained (call it $A$) at $h$: $A \mapsto A(h)$.  Let's call this final map $\mathcal{I}_{h} : \mathcal{L}(E,F) \to F$, $\mathcal{I}_{h} : A \mapsto A(h)$.  Putting it all together, we have $\dot{g}(t) = \mathcal{I}_{h}(Df(x + th))$, which is a composition.
By the chain rule, $\ddot{g}(t) = D\mathcal{I}_{h}(Df(x + th)) \circ D^{2}f(x + th)(h)$.  All that is left is computation of $D\mathcal{I}_{h}$.  This is straightforward, though: $\mathcal{I}_{h}$ is linear so $D\mathcal{I}_{h}(A) = \mathcal{I}_{h}$ for all $A \in \mathcal{L}(E,F)$.  Thus, $\ddot{g}(t) = [D^{2}f(x + th)(h)](h)$.
