Second variation of a functional I am trying to find the second variation of the Hartree energy functional $E_{H} [\rho]$:
$$
\dfrac {\delta^2 E_{H}}{\delta \rho (r)\delta \rho (r')}=\dfrac {\delta^2}{\delta \rho (r)\delta \rho (r')}\iint \dfrac{\rho (r)\rho (r')}{|r-r'|}d^3rd^3r'
$$
but I am a bit confused. Is this just equal to:
$$
\dfrac{1}{|r-r'|}
$$
or is it more complicated and not that trivial? Any hints on how to compute this? 
 A: The rule that defines functional derivatives is
$$\frac{\delta \rho(\mathbf{r})}{\delta \rho(\mathbf{r}')} = \delta(\mathbf{r}-\mathbf{r}').$$
Note that the right hand side is the Dirac delta function. That, and things like obeying the product rule, chain rule, etc. that ordinary derivatives obey.
So, one thing you have making your life difficult is you have given dummy variables inside the integral the same name as variables outside of the integral in your variational derivatives. You can make things clearer by switching out the integration variables to something else (like $\mathbf{x}$ and $\mathbf{x}'$), so
\begin{align}
    \dfrac {\delta^2 E_{H}}{\delta \rho (\mathbf{r})\delta \rho (\mathbf{r}')} &=\dfrac {\delta^2}{\delta \rho (\mathbf{r})\delta \rho (\mathbf{r}')}\iint \dfrac{\rho (\mathbf{x})\rho (\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|}\operatorname{d}^3x\operatorname{d}^3x'\\
   & = \dfrac {\delta}{\delta \rho (\mathbf{r})}\iint \dfrac{\delta (\mathbf{x} - \mathbf{r}')\rho (\mathbf{x}') + \rho(\mathbf{x})\delta(\mathbf{x}'-\mathbf{r}')}{|\mathbf{x}-\mathbf{x}'|}\operatorname{d}^3x\operatorname{d}^3x' \\
   & = \iint \dfrac{\delta (\mathbf{x} - \mathbf{r}')\delta(\mathbf{x}'-\mathbf{r}) + \delta(\mathbf{x}-\mathbf{r})\delta(\mathbf{x}'-\mathbf{r}')}{|\mathbf{x}-\mathbf{x}'|}\operatorname{d}^3x\operatorname{d}^3x'\\
   &\mathrm{etc.}
\end{align}
