I completely understand the proof for the Euler-Lagrange equation for a general function $F(x,y,y')$. However, when I try to use the same proof technique on a function $F(y')$, I run into a curiosity that I can't seem to resolve.
So I want to choose a function $y(x)$ passing through points $x = a$ and $x = b$ which minimizes $$I[y]= \int_a^b F(y') dx.$$
Suppose $y*$ is such a function, then any function $y(x)$ can be written as $$y^*(x)+eV(x),$$
where $e$ is a constant and $V(x)$ is some function of $x$ such that $[y-y^*]/e = V$. Thus the integral $I$ is now a function of $e$, and $I'(0) = 0$. To calculate $I'(e)$, we can take the derivative inside the integral, which is $$\frac{dF}{dy'} \frac{dy'}{de} = \frac{dF}{dy'} V'.$$
Now here is where my confusion is: This integral needs to be $0$ for all $F$ and $V$ as long as $e=0$ ($y=y^*$), so it's fair to say that we can't count on $V$ being a constant (thus $V'$ being 0), and so we must have that $dF/dy' = 0$. However, if we use integration by parts, we get that the integrand is
$$\frac{d}{dx}\frac{dF}{dy'} V,$$
and since we can't count on $V$ being $0$,
$$\frac{d}{dx}\frac{dF}{dy'} = 0,\ \ \text{ or }\frac{dF}{dy'} = C.$$
The latter is also consistent with the Euler-Lagrange equation, so it's definitely correct.
So my question is this: Why can I not set $\frac{dF}{dy'} = 0$ before I do integration by parts, but I can after?
You are using the following lemma:
Lemma: If $f$ is continuous on $[a,b]$ and $\int_a^b f(x) V(x) dx = 0$ for all continuous $V$ so that $V(a) = V(b) = 0$, then $f(x) = 0$.
This is the lemma you used in the argument to conclude that $\frac{d}{dx} \frac{dF}{dy'} = 0$.
However, without using integration by part, you only have
$$\int_a^b g(x) V'(x) dx$$
and all smooth function $V(x)$ with $V(a) = V(b) = 0$. This is not sufficient to infer that $g(x) = 0$. Indeed if $g(x)=C$ is a constant function,
$$\int_a^b CV'(x) dx = C \int_a^b V'(x) dx = C(V(b) - V(a)) = 0,$$
by the fundamental theorem of Calculus. So you can at best conclude $g(x) = C$, instead of $g(x) = 0$.
A proof of the lemma: Let $c\in (a, b)$ so that $f(c)\neq 0$. Without loss of generality, assume that $f(c)>0$. Then there is $\delta>0$, as $f$ is continuous, so that $f(x) >f(c)/2$ for all $|x-c|<\delta$. Now let $V(x)$ be continuous positive function supported inside $(c-\delta, c+\delta)$. Then
$$\int_a^b f(x) V(x) dx >0$$
which is not possible from the assumption. Thus $f(x) = 0$ on $(a, b)$. As $f$ is continuous, it's zero on $[a, b]$.