I ran into the equation below. I'm not familiar with functional derivatives so I'd appreciate if someone could give me an idea of how to solve it and/or a good reference I can use. I appreciate your help!
Let $g=g(v)$, $v=v(x)$ and let $h^{i}=h^{i}(x,v(x),v_{x}(x))$ for $i=1,2,3$. Functions $h^{i}$ are known while function $v(x)$ is unknown. We're looking for a solution of $g$ such that,
$g_{vv}\left(v\right)+h^{1}\left(x,v\left(x\right),v_{x}\left(x\right)\right)g_{v}\left(v\right)+h^{2}\left(x,v\left(x\right),v_{x}\left(x\right)\right)g\left(v\right)+h^{3}\left(x,v\left(x\right),v_{x}\left(x\right)\right)=0$.
Some other conditions are $x\in\left[\underline{x},\infty\right)$, $v\left(\underline{x}\right)>0$ and $v(x)$ is an increasing function of $x$. Also, $\int_{v\left(\underline{x}\right)}^{\infty}g\left(v\right)dv=1$.
While waiting for the OP to reveal more details, I'll try to make sense of what is presented, and later edit if needed. Not an answer, obviously.
We have a single equation for two unknown functions $g(v)$ and $v(x)$:
$$g_{vv}\left(v\right)+h_{1}\left(x,v\left(x\right),v_{x}\left(x\right)\right)g_{v}\left(v\right)+h_{2}\left(x,v\left(x\right),v_{x}\left(x\right)\right)g\left(v\right)+h_{3}\left(x,v\left(x\right),v_{x}\left(x\right)\right)=0$$
First, let us rewrite everything in terms of $x$:
$$g(v(x))=G(x)$$
$$\frac{dg}{dv}=\frac{dG}{dx}\frac{dx}{dv}=\frac{G_x}{v_x}$$
$$\frac{d}{dv}\frac{dg}{dv}=\frac{dx}{dv}\frac{d}{dx} \frac{G_x}{v_x}=\frac{1}{v_x} \left(\frac{G_{xx}}{v_x}-\frac{G_x v_{xx}}{v_x^2} \right)$$
Since each function is of the same variable now, I'll use $'$ for derivatives and rewrite the equation as follows:
$$\frac{G''}{v'^2}-\frac{G' v''}{v'^3}+h_1(x,v,v')\frac{G'}{v'}+h_2(x,v,v')G+h_3(x,v,v')=0 $$
Multiplying by $v'^3$ (from the conditions placed on $v(x)$ we know that $v'>0$):
$$G''v'-G' v''+h_1(x,v,v')G'v'^2+h_2(x,v,v')Gv'^3+h_3(x,v,v')v'^3=0 $$
This is a second order nonlinear ODE (with a single variable $x$), for two unknown functions.
We need to find some pair $G(x),v(x)$ which satisfies the equation and the conditions placed on both functions. Then we have $g(v)$ in parametric form.
I'm sure there's a lot of different solutions here.
In a more common case, we would have another ODE with the two functions, then we would be able to find the general solution (provided we even know it for this particular kind of nonlinear equations).
This is similar to a case of a, say, algebraic equation $P(y,x,t)=0$ where we were asked to find $y(x)$. A lot of pairs $x(t),y(x)$ might be solutions.
To the question in the title, I would likely say that this is not a functional differential equation, at least not any kind that I heard of. Usually functional ODEs are known to have terms with changed argument of some kind, but the change is known. For example $y'(x)=y(x-1)$ or $y'(x)=y(2x)$ or something more complicated.