This is a common question in calculus of variations, and the issue comes down to not understanding the notation used. Here, $L$ is a function of three variables, $L: \Bbb{R}^3 \to \Bbb{R}$, defined as
\begin{align}
L(\xi,\eta,\zeta) &= \sqrt{1+ \zeta^2}
\end{align}
What the Euler-Lagrange equations are saying is that a (sufficiently nice) function $y: [a,b] \to \Bbb{R}$ makes the "action" stationary if and only if for every $x \in [a,b]$, we have
\begin{align}
\dfrac{\partial L}{\partial \eta}\bigg|_{(x,y(x), y'(x))} - \dfrac{d}{dx}\left(\dfrac{\partial L}{\partial \zeta}\bigg|_{(x,y(x), y'(x))} \right) &= 0
\end{align}
Or expressed in slightly different notation (which I think is the most precise way of expressing it)
\begin{align}
(\partial_2L)_{(x,y(x), y'(x))} - \dfrac{d}{ds}\bigg|_{s=x} \bigg((\partial_3L)_{(s,y(s), y'(s))} \bigg) &= 0
\end{align}
Here is what the notation means. $L$ is a function of $3$ variables, so it has $3$ partial derivatives $\partial_1L, \partial_2L$ and $\partial_3L$. These are again functions $\Bbb{R}^3 \to \Bbb{R}$. So, the meaning of $(\partial_2L)_{(x,y(x),y'(x))}$ means that real number which is obtained when you apply the function $\partial_2L$ to the input $(x,y(x), y'(x)) \in \Bbb{R}^3$.
Similarly, $s\mapsto (\partial_3L)_{(s,y(s), y'(s))}$ is a function $\Bbb{R} \to \Bbb{R}$, and in the second term of the Euler-Lagrange equations, we want to consider the derivative of this single-variable function at the point $x$. Ultimately the key thing to remember with Calculus of variations and Euler-Lagrange equations is that you calculate partial derivatives FIRST, and then evaluate
For some reason people don't want to make this point explicit when teaching Euler-Lagrange equations and calculus of variations, which leads to questions like "how are the function and the derivative independent variables"? The answer is of course a function and its derivative are not independent... that would be extremely weird. The point is we are plugging in the curve only after performing the derivatives.
In the above example, $L(\xi,\eta, \zeta) = \sqrt{1+\zeta^2}$. So, what is $\dfrac{\partial L}{\partial \zeta}$? Simple:
\begin{align}
\dfrac{\partial L}{\partial \zeta} \bigg|_{(\xi,\eta,\zeta)} &= \dfrac{\zeta}{\sqrt{1+ \zeta^2}},
\end{align}
or in more precise notation,
\begin{align}
(\partial_3L)_{(\xi, \eta, \zeta)} &= \dfrac{\zeta}{\sqrt{1+ \zeta^2}}.
\end{align}
i.e the RHS is the value of the third partial derivative when evaluated at the point $(\xi,\eta,\zeta) \in \Bbb{R}^3$. For example,
\begin{align}
(\partial_3L)_{(1, 4, 8)} &= \dfrac{8}{\sqrt{1+ 8^2}} = \dfrac{8}{\sqrt{65}}
\end{align}
Now, suppose you have a function $y: [a,b] \to \Bbb{R}$. What is the value of the above partial derivative when evaluated at the point $(x,y(x), y'(x))$? Very simple:
\begin{align}
(\partial_3L)_{(x,y(x), y'(x))} &= \dfrac{y'(x)}{\sqrt{1+ (y'(x))^2}}
\end{align}
So, if for example $y(x) = x^2$ then $y'(x) = 2x$. So,
\begin{align}
(\partial_3L)_{(x,y(x), y'(x))} &= (\partial_3L)_{(x,x^2, 2x)} = \dfrac{2x}{\sqrt{1+ (2x)^2}} = \dfrac{2x}{\sqrt{1+4x^2}}
\end{align}
