Q:

Functional Derivative of Convolution

How to carry out the following functional derivative? 
$$\frac{\delta F}{\delta n(r)}$$ where $$F=\int dr n(r) \int C(|r-r'|) n(r') dr'$$
is it simply:
$$2 \int dr' C(|r-r'|) n(r')$$? 

A:

Notice that $F$ is essentially a quadratic form; i.e. if it were matrices then you would have (in summation notation): $$F = x_i C_{ij} x_j.$$ Then you would use the fact that $\frac{\partial x_i}{\partial x_j} = \delta_{ij}$ to get
\begin{align}
\frac{\partial F}{\partial x_k} &= \delta_{ik} A_{ij} x_{j} + x_i A_{ij} \delta_{jk} \
  &= 2 A_ik x_k
\end{align}
if $A_{ij} = A_{ij}$ i.e. it is symmetric. 
Here, we use a similar fact: $$\frac{\delta n(x)}{\delta n(y)} = \delta(x-y)$$ where $\delta$ this time is the Dirac distribution.  Your "matrix" in the middle is obviously symmetric, so your proposed answer is correct.