Suppose I take a space $M$ of smooth functions $\rho:X\rightarrow\mathbb{R}$, and I give $M$ some manifold structure (perhaps as a Banach or Hilbert manifold?).
Then if I have a function $F:M\rightarrow \mathbb{R}$, the functional differential $\delta F$ takes two arguments from $M$, $\rho$ and $\phi$, such that
$$ \delta F(\rho,\phi) = \lim_{\epsilon\rightarrow 0}\frac{F(\rho+\epsilon\phi)- F(\rho)}{\epsilon}$$
But since $M$ has an additive structure, I could instead define a path $\gamma_{\rho,\phi}:[-1,1]\rightarrow M$ such that $\gamma_{\rho,\phi}(t)=\rho+t\phi$. Then if I define $\partial_{\phi,\rho}:= \gamma_{\rho,\phi}'(0) \in T_\rho M$, it's true that
$$ dF_\rho(\partial_{\phi,\rho}) = \partial_{\phi,\rho}(F) = \lim_{\epsilon\rightarrow 0}\frac{F(\rho+\epsilon\phi)- F(\rho)}{\epsilon}=\delta F(\rho,\phi)$$
where $dF$ is the differential of $F$.
If the tangent vectors $\partial_{\phi,\rho}$ span the tangent space of $M$ at every $\rho$, then I can conclude that $dF_{\rho} = \delta F(\rho, \cdot)$ (and maybe they just need to span a dense subset since $\delta F$ and $dF$ both seem to be continuous). That is, functional differentials are just regular differentials, but for a special class of manifolds. Is this true?
The functional derivative seems to be special, though. In general $dF_\rho$ is a linear function on $T_\rho M$, so unless $T_\rho M$ consists of smooth functions, it won't correspond to a measure. Here I've identified $M$ with its tangent space so elements of $T_\rho M$ can be regarded as smooth functions.