The quadratic Lagrangian is given by

\[ L = g_{\mu\nu}\dot{X}^{\mu}\dot{X}^{\nu}. \]

The Euler–Lagrange equation is

\[ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{X}^{\lambda}}\right) = \frac{\partial L}{\partial X^{\lambda}}. \]

We first evaluate \(\frac{\partial L}{\partial \dot{X}^{\lambda}}\):

\[ \begin{aligned} \frac{\partial L}{\partial \dot{X}^{\lambda}} &= \underbrace{\frac{d g_{\mu\nu}}{d\dot{X}^{\lambda}}}_{0} \dot{X}^{\mu}\dot{X}^{\nu} + g_{\mu\nu}\frac{d\dot{X}^{\mu}}{d\dot{X}^{\lambda}}\dot{X}^{\nu} + g_{\mu\nu}\dot{X}^{\mu}\frac{d\dot{X}^{\nu}}{d\dot{X}^{\lambda}} && \text{(product rule)} \\ &= g_{\mu\nu}\delta^{\mu}_{\lambda}\dot{X}^{\nu} + g_{\mu\nu}\dot{X}^{\mu}\delta^{\nu}_{\lambda} && \left(\frac{d\dot{X}^{a}}{d\dot{X}^{b}}=\delta^{a}_{b}\right) \\ &= g_{\lambda\nu}\dot{X}^{\nu}+g_{\mu\lambda}\dot{X}^{\mu} && \text{(Kronecker delta contraction)} \\ &= 2g_{\lambda\nu}\dot{X}^{\nu} && \left(g_{\lambda\nu}=g_{\nu\lambda}\right). \end{aligned} \]

Now we evaluate \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{X}^{\lambda}}\right)\):

\[ \begin{aligned} \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{X}^{\lambda}}\right) &= \frac{d}{dt}\left(2g_{\lambda\nu}\dot{X}^{\nu}\right) \\ &= 2\frac{dX^{\alpha}}{dt}\frac{d}{dX^{\alpha}}\left(g_{\lambda\nu}\right) \dot{X}^{\nu}+2g_{\lambda\nu}\ddot{X}^{\nu} && \text{(product and chain rules)} \\ &= 2\partial_{\alpha}g_{\lambda\nu}\dot{X}^{\nu}\dot{X}^{\alpha} +2g_{\lambda\nu}\ddot{X}^{\nu}. \end{aligned} \]

Before we evaluate \(\frac{\partial L}{\partial X^{\lambda}}\), we first show that \(\partial_{\lambda}\dot{X}^{\mu}=0\):

\[ \begin{aligned} \frac{\partial}{\partial X^{\lambda}} \left(\frac{\partial X^{\mu}}{\partial t}\right) &= \frac{\partial^{2}X^{\mu}}{\partial X^{\lambda}\,\partial t} && \text{(definition of mixed derivative)} \\ &= \frac{\partial}{\partial t}\left(\delta^{\mu}_{\lambda}\right) && \left(\frac{\partial X^{\mu}}{\partial X^{\lambda}}=\delta^{\mu}_{\lambda}\right) \\ &= 0 && \text{(the Kronecker delta is constant)}. \end{aligned} \]

Now we evaluate \(\frac{\partial L}{\partial X^{\lambda}}\):

\[ \begin{aligned} \frac{\partial L}{\partial X^{\lambda}} &= \left(\partial_{\lambda}g_{\mu\nu}\right)\dot{X}^{\mu}\dot{X}^{\nu} +g_{\mu\nu}\left(\partial_{\lambda}\dot{X}^{\mu}\right)\dot{X}^{\nu} +g_{\mu\nu}\dot{X}^{\mu}\left(\partial_{\lambda}\dot{X}^{\nu}\right) && \text{(product rule)} \\ &= \left(\partial_{\lambda}g_{\mu\nu}\right)\dot{X}^{\mu}\dot{X}^{\nu} && \text{(based on the result above)}. \end{aligned} \]

To summarize, we have

\[ \begin{aligned} \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{X}^{\lambda}}\right) &=2\partial_{\alpha}g_{\lambda\nu}\dot{X}^{\nu}\dot{X}^{\alpha} +2g_{\lambda\nu}\ddot{X}^{\nu}, \\ \frac{\partial L}{\partial X^{\lambda}} &=\left(\partial_{\lambda}g_{\mu\nu}\right)\dot{X}^{\mu}\dot{X}^{\nu}. \end{aligned} \]

From the Euler–Lagrange equation, we equate them:

\[ \begin{aligned} 2\left(\partial_{\alpha}g_{\lambda\nu}\dot{X}^{\alpha}\dot{X}^{\nu} +g_{\lambda\nu}\ddot{X}^{\nu}\right) &=\partial_{\lambda}g_{\mu\nu}\dot{X}^{\mu}\dot{X}^{\nu} && \text{(Euler--Lagrange equation)} \\ 2g_{\lambda\nu}\ddot{X}^{\nu} &=\partial_{\lambda}g_{\mu\nu}\dot{X}^{\mu}\dot{X}^{\nu} -2\partial_{\alpha}g_{\lambda\nu}\dot{X}^{\alpha}\dot{X}^{\nu} && \text{(isolate the acceleration term)} \\ &=\partial_{\lambda}g_{\mu\nu}\dot{X}^{\mu}\dot{X}^{\nu} -2\cdot\frac{1}{2}\left( \partial_{\alpha}g_{\lambda\nu}\dot{X}^{\alpha}\dot{X}^{\nu} +\partial_{\nu}g_{\lambda\alpha}\dot{X}^{\nu}\dot{X}^{\alpha} \right) && \text{(symmetrize in }\alpha\text{ and }\nu\text{)} \\ \frac{1}{2}g^{\lambda\rho}\cdot 2g_{\lambda\nu}\ddot{X}^{\nu} &=\frac{1}{2}g^{\lambda\rho}\left( \partial_{\lambda}g_{\mu\nu}\dot{X}^{\mu}\dot{X}^{\nu} -\partial_{\alpha}g_{\lambda\nu}\dot{X}^{\alpha}\dot{X}^{\nu} -\partial_{\nu}g_{\lambda\alpha}\dot{X}^{\nu}\dot{X}^{\alpha} \right) && \text{(multiply both sides by }\tfrac{1}{2}g^{\lambda\rho}\text{)} \\ \ddot{X}^{\rho} &=\frac{1}{2}g^{\lambda\rho}\left( \partial_{\lambda}g_{\mu\nu}\dot{X}^{\mu}\dot{X}^{\nu} -\partial_{\alpha}g_{\lambda\nu}\dot{X}^{\alpha}\dot{X}^{\nu} -\partial_{\nu}g_{\lambda\alpha}\dot{X}^{\nu}\dot{X}^{\alpha} \right) && \text{(simplify)} \\ \ddot{X}^{\rho} &=\frac{1}{2}g^{\lambda\rho}\left( \partial_{\lambda}g_{\alpha\nu} -\partial_{\alpha}g_{\lambda\nu} -\partial_{\nu}g_{\lambda\alpha} \right)\dot{X}^{\alpha}\dot{X}^{\nu} && \text{(rename indices and factor)}. \end{aligned} \]

Now recall the derivation of the Christoffel symbols from the metric:

\[ -\Gamma^{\rho}_{\alpha\nu} =\frac{1}{2}g^{\lambda\rho}\left( \partial_{\lambda}g_{\alpha\nu} -\partial_{\alpha}g_{\lambda\nu} -\partial_{\nu}g_{\lambda\alpha} \right). \]

Thus we have

\[ \ddot{X}^{\rho}=-\Gamma^{\rho}_{\alpha\nu}\dot{X}^{\alpha}\dot{X}^{\nu}, \]

which are the geodesic equations satisfying \(\nabla_{\rho}\dot{X}=0\).