The expression
$\delta F[\rho,\phi] := \frac{dF[\rho(x) + \epsilon \phi(x)]}{d\epsilon}\Big|_{\epsilon=0},$
when defined, is a functional of $\rho$ and $\phi.$ The dependency on $\rho$ is usually non-linear, while the dependency on $\phi$ is usually linear.
If the expression is restricted to $\phi \in C_c^\infty(\mathbb R^n)$ and the dependency on $\phi$ is linear, then the mapping $\phi \mapsto \delta F[\rho,\phi]$ is usually a distribution. Often this distribution can be identified with a function.
Thus, $\delta F[\rho,\phi]$ is a functional, usually a distribution, and often a function.
Often we have $F[\rho] = \int L(x, \rho(x), \rho'(x)) \, dx$ for some Lagrangian $L.$ Then, if $\phi$ vanishes on the boundary of the domain,
$$
\delta F[\rho,\phi] = \int \left( \frac{\partial L}{\partial \rho} \phi(x) + \frac{\partial L}{\partial \rho'} \phi'(x) \right) dx
= \int \left( \frac{\partial L}{\partial \rho} - \frac{d}{dx}\frac{\partial L}{\partial \rho'} \right) \phi(x) \, dx.
$$
In this case, $\delta F[\rho,\phi]$ is given by an integral of a function (the parenthesis) times $\phi.$ Thus this falls into the case "Often this distribution can be identified with a function".
Yes, this is a simple case. The variational differential is
$$
\delta F[\rho] = \int f'(\rho(r)) \, \delta\rho(r) \, dr.
$$
Best Answer
Wikipedia defines the functional derivative through the formula (see this section) $$\left.\frac{\mathrm{d}}{\mathrm{d}\epsilon} F[\rho + \epsilon \phi]\right|_{\epsilon=0} =: \int \frac{\delta F}{\delta \rho(x)} \phi(x) \mathrm{d}x.$$
Suppose now we want to perform a change of coordinates $x \to y$. Then we have $$\left.\frac{\mathrm{d}}{\mathrm{d}\epsilon} F[\rho + \epsilon \phi]\right|_{\epsilon=0} =: \int \frac{\delta F}{\delta \rho(x(y))} \phi(x(y)) \left|\frac{\mathrm{d}x}{\mathrm{d}y}\right| \mathrm{d}y,$$ and we see a Jacobian factor has appeared. Of course, $\phi(x(y)) = \phi(x)$, and hence we must conclude that $$\frac{\delta F}{\delta \rho(y)} = \left|\frac{\mathrm{d}x}{\mathrm{d}y}\right| \frac{\delta F}{\delta \rho(x)}.$$