Are functional differentials just regular differentials on the manifold of functionals?

Suppose I take a space $M$ of smooth functions $rho:Xrightarrowmathbb{R}$, and I give $M$ some manifold structure (perhaps as a Banach or Hilbert manifold?).

Then if I have a function $F:Mrightarrow mathbb{R}$, the functional differential $delta F$ takes two arguments from $M$, $rho$ and $phi$, such that

$$ delta F(rho,phi) = lim_{epsilonrightarrow 0}frac{F(rho+epsilonphi)- F(rho)}{epsilon}$$

But since $M$ has an additive structure, I could instead define a path $gamma_{rho,phi}:[-1,1]rightarrow M$ such that $gamma_{rho,phi}(t)=rho+tphi$. Then if I define $partial_{phi,rho}:= gamma_{rho,phi}'(0) in T_rho M$, it’s true that

$$ dF_rho(partial_{phi,rho}) = partial_{phi,rho}(F) = lim_{epsilonrightarrow 0}frac{F(rho+epsilonphi)- F(rho)}{epsilon}=delta F(rho,phi)$$

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.