This is the expression for the energy of the hydrogen atom, which can be expressed as the following integral
$$ E = \int \psi^* \cdot \hat{H} \cdot \psi d \tau \\ \hat{H}=-\frac{\hbar^2}{2 m_e r^2} \frac{d}{d r}\left(r^2 \frac{d}{d r}\right)-\frac{e^2}{4 \pi \varepsilon_0} $$
where $\hbar$, $m_e$ and $\epsilon$ are known variables and $\psi^*$ is the conjugate transpose of $\psi$. We can think of this as $f$ in the Euler-Lagrangian quantity:
$$ \psi^* \cdot \hat{H} \cdot \psi $$
If I plug $\hat{H}$ in, I get:
$$ f\left(r,\psi, \frac{d\psi}{dr}\right)=\psi^* \cdot \frac{1}{r^2} \frac{d}{d r}\left(r^2 \frac{d \psi}{d r}\right)-\psi^*\frac{e^2}{4 \pi \varepsilon r} \cdot \psi $$
And then I can plug $f$ into the Euler-Lagrange equation,Now the question is, I don't know how to plug it in, because it has a second derivative and $\psi^*$, right
According to the comments, the Euler-Lagrange formula of the second derivative is required, I asked ChatGPT
$$ \frac{\partial L}{\partial \psi}-\frac{d}{d r}\left(\frac{\partial L}{\partial\left(\partial_r \psi\right)}\right)+\frac{d^2}{d r^2}\left(\frac{\partial L}{\partial\left(\partial_r^2 \psi\right)}\right) = 0 $$
I'm not going to deal with $psi^*$, but I'm going to assume $psi^*$ is a real function,So here's how I do it
Am I doing the right things?
$$L_{\Psi+\epsilon} = (\frac{\bar \Psi + \bar \epsilon}{r^2})\frac{d}{dr}(r^2\frac{d(\Psi + \epsilon)}{dt}) - \frac{e^2(\bar \Psi + \bar \epsilon)}{4\pi\epsilon_0r}(\Psi + \epsilon)$$
$$L_{\Psi} = \frac{\bar \Psi}{r^2}\frac{d}{dr}(r^2\frac{d\Psi}{dt}) - \frac{e^2}{4\pi\epsilon_0r}\bar \Psi\Psi$$
Then,
$$L_{\Psi+\epsilon} - L_{\Psi} = \bar \epsilon\frac{d^2\Psi}{dr^2} + \frac{2}{r}\bar \epsilon \frac{d\Psi}{dr} + \bar \epsilon\frac{d^2\epsilon}{dr^2} + \frac{2}{r}\bar\Psi \frac{d\epsilon}{dr} + \frac{2}{r}\bar\epsilon\frac{d\epsilon}{dr}-\frac{e^2}{4\pi\epsilon_0r}\epsilon$$
$$=\bar \epsilon (\frac{d^2\Psi}{dr^2} + \frac{2}{r}\frac{d\Psi}{dr} - \frac{e^2}{4\pi\epsilon_0r}) + \bar \epsilon \frac{d^2\epsilon}{dr^2} + \Psi\frac{d^2\epsilon}{dr^2} + \frac{2}{r}\bar\Psi\frac{d\epsilon}{dr}$$
Suppose that all terms involving $\frac{d\epsilon}{dr}$ and higher derivatives vanish, then
$$\frac{d^2\Psi}{dr^2} + \frac{2}{r}\frac{d\Psi}{dr} - \frac{e^2}{4\pi\epsilon_0r} = 0$$
is the Euler-Lagrange equation, as it will make the first term vanish as well, giving $L_{\Psi+\epsilon} - L_{\Psi}=0$ which is an extremum of L over arbitrary functionals.