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What is the functional derivative of a functional $F$ that is expressed as a volume integral over a region $Omegasubsetmathbb R^3$ plus a surface integral over the boundary $partialOmega$?
An example for such a functional is
$$
F[c] = int_Omega f(c, nabla c) , mathrm{d}V + oint_{partialOmega} g(c) , mathrm{d} S
;.
$$
I think that inside the domain the functional derivative reads
$$
frac{delta F}{delta c} = frac{partial f}{partial c}
- nabla frac{partial f}{partial (nabla c)}
;,
$$
but I do not know how to deal with the boundary. I'm not even sure whether the problem is well-posed (even assuming reasonably nice properties of $Omega$, $f$, and $g$).

My more general question therefore is how one deals with functionals of the aforementioned structure.

Generally speaking, the boundary terms in a functional only contributes to the boundary conditions, and would not lead changes to the variational derivative.

For your example, repeat the derivation of the classical Euler-Lagrange equation, and
begin{align}
delta F&=int_{Omega}delta f(c,nabla c),{rm d}V+int_{partialOmega}delta g(c),{rm d}S\
&=int_{Omega}left(frac{partial f}{partial c}delta c+frac{partial f}{partialleft(nabla cright)}cdotnabladelta cright){rm d}V+int_{partialOmega}g'(c)delta c,{rm d}S\
&=int_{Omega}left(frac{partial f}{partial c}delta c+nablacdotleft(frac{partial f}{partialleft(nabla cright)},delta cright)-nablacdotleft(frac{partial f}{partialleft(nabla cright)}right)delta cright){rm d}V+int_{partialOmega}g'(c)delta c,{rm d}S\
&=int_{Omega}left(frac{partial f}{partial c}-nablacdotleft(frac{partial f}{partialleft(nabla cright)}right)right)delta c,{rm d}V+int_{partialOmega}left(frac{partial f}{partialleft(nabla cright)}cdot n+g'(c)right)delta c,{rm d}S,
end{align}
where $n$ denotes the outward unit normal vector on $partialOmega$.

Therefore, the $c$ that minimizes $F$ must satisfy
begin{align}
frac{partial f}{partial c}-nablacdotleft(frac{partial f}{partialleft(nabla cright)}right)&=0quadtext{in }Omega,\
frac{partial f}{partialleft(nabla cright)}cdot n+g'(c)&=0quadtext{on }partialOmega
end{align}
due to the arbitrariness of $delta c$.

The problem can be solved by calculating the first variation of the functional from scratch.
Here, the first variation reads $delta F
= lim_{epsilon rightarrow 0}(F[c + delta c] - F[c])$ with $delta c =epsilon phi$ and $phi$ is a arbitrary function.
We thus have
begin{align}
F[c + epsilon phi] &=
int_Omega f(c + epsilonphi, nabla c + epsilon nabla phi) mathrm{d}^3 r
+
oint_{partialOmega} g(c + epsilonphi) mathrm{d}^2 r
end{align}
which can be expanded to first order in $epsilon$,
begin{align}
F[c + epsilon phi] &=
int_Omegaleft[
f(c, nabla c)
+ frac{partial f}{partial c}epsilonphi
+ frac{partial f}{partial (nabla c)}epsilon nabla phi
right] mathrm{d}^3 r
+
oint_{partialOmega} left[
g(c)
+ g'(c)epsilonphi)
right] mathrm{d}^2 r
;.
end{align}
We thus obtain for the first variation
begin{align}
delta F &=
int_Omegaleft[
frac{partial f}{partial c}delta c
+ frac{partial f}{partial (nabla c)} nabla delta c
right] mathrm{d}^3 r +
oint_{partialOmega} !!! g'(c)delta c , mathrm{d}^2 r
end{align}
Using integration by parts
begin{align}
delta F &=
int_Omegaleft[
frac{partial f}{partial c}delta c
- delta cnablafrac{partial f}{partial(nabla c)}
right] mathrm{d}^3 r
+
oint_{partialOmega}left[
frac{partial f}{partial (partial_alpha c)} n_alpha delta c
+ g'(c)delta c
right] mathrm{d}^2 r
;,
end{align}
where $n_alpha$ is the normal vector of the boundary.
In particular, the associated Euler-Lagrange equations read
begin{align}
0 &= frac{partial f}{partial c} - nablafrac{partial f}{partial (nabla c)}
\
0 &= frac{partial f}{partial(partial_alpha c)} n_alpha + g'(c)
label{eqn:stationary_point_boundary}
end{align}