In this example of using the Euler-Lagrange equation, I am struggling to understand as to what the term being differentiate with respect to $x$ in line $(2.30)$ came from (it's not one of the partial derivatives of $F$ with respect to $y$ or $y'$). I am thinking that it might be from the equation: $$F_{y'y'}y'' = F_y - F_{y'y}y' - F_{y'x}$$ where $F_{y'x} = 0$. But the term I am confused about isn't even equal to $F_{y'}$.
On
Note that 2.29 says that $yy'' = 1+(y')^2$. Rewriting, we get $1+(y')^2 - yy'' = 0$, which implies $$\frac{y'(1+(y')^2 - yy'')}{(1+(y')^2)^{3/2}} = 0$$where the left-hand side happens to be equal to $$\frac d{dx}\left(\frac{y}{\sqrt{1+(y')^2}}\right)$$This function that we differentiate is constructed explicitly to make this work. It is, so to speak, pulled out of a magician's hat in order to get a single derivative on one side so that we may integrate. It is a common enough technique when solving differential equations. When you see terms like $yy''$ and $(y')^2$ flying around, then at least to me it smells of product or quotient rule, and I would try to rearrange it into something that works, but this is so complicated that I think it would've taken me a long time to find.
For any magic there is a simple scientific explanation. In this case it is called Beltrami identity. Since the function $F$ does not depend explicitly on $x$, the Euler-Lagrange equation is equivalent (modulo to constants $y'=0$) to the following equation $$ F-y'F_{y'}=C, $$ which after the differentiation and simplification becomes $$ \frac{y}{\sqrt{1+y'^2}}=C. $$ Here we recognize our old fellow in the LHS.