This post was inspired by concepts in physics, but it's math oriented.
We can use Euler-Lagrange equation to find a function $y(x)$ which produces the stationary action:
$$\int_{x_1}^{x_2} \mathcal{L}(y, \frac{dy}{dx}, x) dx \implies \frac{d}{dx}\frac{\partial \mathcal{L}}{\partial y'} – \frac{\partial\mathcal{L}}{\partial y} = 0.\tag{1}$$
There is a field equivalent for a function $\psi(\vec{x})$ that takes in $n$ scalars and outputs a scalar:
$$\int_V \mathcal{L}(\psi(\vec{x}), \nabla \psi(\vec{x}), \vec{x}) d^n\vec{x} \implies \sum_{i=1}^n\frac{d}{dx^i}\frac{\partial \mathcal{L}}{\partial \frac{\partial\psi}{\partial x^i}} – \frac{\partial\mathcal{L}}{\partial \psi} = 0.\tag{2}$$
Is there a functional version? For a functional of a field $F[\psi]$, which takes in a function and gives a scalar, would the action and EL be(?):
$$\int \mathcal{L}[F, \frac{\delta F}{\delta \psi}, \psi] \mathcal{D}(\psi) \stackrel{?}{\implies} \int\frac{\delta}{\delta\psi(x)} \frac{\delta \mathcal{L}}{\delta \frac{\delta F}{\delta \psi(x)}}d^nx-\frac{\delta \mathcal{L}}{\delta F}=0.\tag{3}$$
Best Answer
Since the path/functional integral is generically a mathematically ill-defined construct, it would not be easy to rigorously develop the correspondent calculus of variations, but heuristically OP's functional Euler-Lagrange (EL) equation (3) makes sense.
In the context of physics ${\cal L}=\exp(\frac{i}{\hbar}F[\psi])$ in eq. (3) could be the Boltzmann factor, and $F[\psi]$ the action functional. Usually the action is given in physics applications, but here we are apparently looking at a space of theories/actions, and are trying to determine which theory/action is a stationary point for the path integral. The Boltzmann factor ${\cal L}$ does not depend on the functional derivative $\frac{\delta F}{\delta \psi}$ so the functional Euler-Lagrange (EL) equation (3) here simplifies.