This question is specifically about deriving the Beltrami identity.
Just to give this question context I provide an example of a problem that is solved with Calculus of Variations: find the shape of a soap film that stretches between two coaxial rings.
For the surface area the expression to be integrated from start point to end point:
$$ F = 2 \pi \int_{x_0}^{x_1} y \ \sqrt{1 + (y')^2} \ dx \tag{1} $$
For the purpose of finding the function that minimizes that surface area the Euler-Lagrange equation is applied.
As we know, since the value of F does not depend directly on the x-coordinate the Beltrami identity is applicable.
Comparison of the EL-equation and the Beltrami identity:
Euler-Lagrange:
$$ \frac{\partial F}{\partial y} - \frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right) = 0 \tag{2} $$
Beltrami:
$$ F - y' \frac{\partial F}{\partial y'} = C \tag{3} $$
We see that the process of conversion from EL-eq. to Beltrami consists of integration with respect to the y-coordinate
For the first term:
$$ \int \frac{\partial F}{\partial y} dy = F + C \tag{4} $$
with $C$ an arbitrary integration constant.
Question:
Is there a transparent way to evaluate the same integral for the second term?
$$ \int \frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right)dy + C \quad = ? = \quad \frac{dy}{dx} \frac{\partial F}{\partial y'} \tag{5} $$
The usual way to arrive at the Beltrami identity is quite a Rube Goldberg derivation.
The thing is: showing that (5) is indeed correct is worthwhile only if it can be done in a way that is more accessible than the usual way of obtaining the Beltrami identity.
The relation between the Euler-Lagrange equation and the Beltrami identity is: the Beltrami identity is the Euler-Lagrange equation with a differentiation with respect to $x$ backed out.
In order to back out a differentiation (here a differentiation with respect to $x$), one must arrive at an expression of the following form:
$$ \frac{dF}{dx} = \frac{dG}{dx} \tag{1} $$
Rearrange:
$$ \frac{dF}{dx} - \frac{dG}{dx} = 0 \tag{2} $$
Differentiation is distributive, so we can convert to:
$$ \frac{d}{dx} (F - G) = 0 \tag{3} $$
At this point we can forego the differentiation $\tfrac{d}{dx}$ and state:
$$ (F - G) = C \tag{4} $$
Where $C$ is a constant, to be determined.
In further preparation:
The product rule:
$$ \frac{d(f(x)g(x))}{dx} = \frac{d\ f(x)}{dx} \ g(x) + f(x) \ \frac{d\ g(x)}{dx} \tag{5} $$
Here the product rule will be used in reverse: it will be used to collapse two terms into one.
$$ \frac{d\ f(x)}{dx} \ g(x) + f(x) \ \frac{d\ g(x)}{dx} = \frac{d(f(x)g(x))}{dx} \tag{6} $$
That completes the preparations.
The general expression for derivative of an expression $F$ with respect to $x$
$$ \frac{dF}{dx} = \frac{\partial F}{\partial x}\frac{dx}{dx} + \frac{\partial F}{\partial y}\frac{dy}{dx} + \frac{\partial F}{\partial y'}\frac{dy'}{dx} \tag{7} $$
Omit the term for partial derivative with respect to $x$:
$$ \frac{dF}{dx} = \frac{\partial F}{\partial y}\frac{dy}{dx} + \frac{\partial F}{\partial y'}\frac{dy'}{dx} \tag{8} $$
In order to go from (8) to the Beltrami identity we need to accomplish the following three objectives:
$$ \frac{\partial F}{\partial y} = \frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right) \tag{9} $$
(9) is the Euler-Lagrange equation, we use it to substitute the factor $\tfrac{\partial F}{\partial y}$ on the right hand side of (8).
The substitution gives (10).
$$ \frac{dF}{dx} = y' \frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right) + y'' \frac{\partial F}{\partial y'} \tag{10} $$
The substitution has accomplished all three of the objectives:
The right hand side of (10) has the same pattern as the left hand side of (6), so those two terms fold into one.
$$ \frac{dF}{dx} = \frac{d}{dx} \left( y' \frac{\partial F}{\partial y'} \right) \tag{11} $$
(11) has the same form as (1), so we can back out a differentiation with respect to $x$
$$ \frac{d}{dx} \left(F - y' \frac{\partial F}{\partial y'} \right) = 0 \tag{12} $$
$$ F - y' \frac{\partial F}{\partial y'} = C \tag{13} $$
The key point:
The relation between the Euler-Lagrange equation and the Beltrami identity is one of integration-differentiation. The difference is differentiation with respect to $x$.