Euler-Lagrange equation for functionals

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(?):

$$\int \mathcal{L}[F, \frac{\delta F}{\delta \psi}, \psi] \mathcal{D}(\psi) \stackrel{?}{\implies} \int\frac{\delta}{\delta\psi(x)} \frac{\delta \mathcal{L}}{\delta \frac{\delta F}{\delta \psi(x)}}d^nx-\frac{\delta \mathcal{L}}{\delta F}=0.\tag{3}$$

2. 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.