>Now, I'd like to know how to define higher order derivatives of functional derivatives. In other words, suppose the Fréchet derivative of $f$ at $\alpha$, $Df(\alpha)$ is Fréchet differentiable at $\beta\in F$, is it possible to define $$
\dfrac{\delta^{2}f}{\delta \beta\delta\alpha}?
$$

Yes, it is possible to define higher order Fréchet derivatives directly as bilinear functionals $\mathscr{L}_2(F,\Bbb R)$. This is shown by Ambrosetti and Prodi ([1], pp. 23-29), for example, and while leaving to them the details, we can simply say that $f:F\to\Bbb R$ is twice *Fréchet differentiable* at $\alpha\in F$ iff

* $F$ is one time Fréchet differentiable and
* $Df[\alpha+\beta](\gamma)-Df[\alpha](\gamma)= \mathfrak{B}[\alpha](\beta,\gamma) + o(\beta)$
where $o(\beta)/\Vert\beta\Vert_F\to 0$ as $\Vert\beta\Vert_F\to 0$ and $\mathfrak{B}\in\mathscr{L}\big(F,\mathscr{L}(F,\Bbb R)\big)\simeq\mathscr{L}_2(F,\Bbb R)$.

This can bee seen also by using the definition of Fréchet differentiability you give, but it seems to me that, when dealing with functional derivatives of order greater than one, it overshadows the intrinsic clarity and familiarity of the concept: 
$$
\begin{split}
Df[\alpha+\varepsilon\beta](\gamma)-Df[\alpha](\gamma) &= \left\langle \frac{\delta f}{\delta \alpha}(\alpha+\beta),\gamma\right\rangle -\left\langle \frac{\delta f}{\delta \alpha} (\alpha),\gamma\right\rangle\\
& = \left\langle\frac{\delta f}{\delta \alpha}(\alpha+\beta)- \frac{\delta f}{\delta \alpha} (\alpha),\gamma\right\rangle\\
&=\left\langle\mathfrak{L}[\alpha](\beta)+ o(\beta),\gamma\right\rangle
\end{split}
$$

**Bibliography**

[1]<cite authors="Ambrosetti, A.; Prodi, G."> Ambrosetti, Antonio; Prodi, Giovanni, _A primer of nonlinear analysis_,  Cambridge Studies in Advanced Mathematics, 34. Cambridge: Cambridge University Press, pp. viii+171 (1993), ISBN: 0-521-37390-5, [MR1225101](http://www.ams.org/mathscinet-getitem?mr=MR1225101), [ZBL0781.47046](https://zbmath.org/?q=an:0781.47046).</cite>