Let us do the RHS first. This just gives us a derivative on the metric:
$$\frac{\partial L}{\partial x^\lambda}=\frac{1}{2}\partial_\lambda g_{\mu\nu}\dot x^\mu\dot x^\nu$$
The first derivative on the LHS is essentially a derivative of a square, thus
$$\frac{\partial L}{\partial \dot x^\lambda}=g_{\mu\lambda}(x(\lambda))\dot x^\mu$$
where we have made the dependence of $g$ on $\lambda$ clear for the next step. Now we differentiate with respect to the curve parameter:
$$\frac{\mathrm{d}}{\mathrm{d}\lambda}[g_{\mu\lambda}(x(\lambda))\dot x^\mu]=\partial_\nu g_{\mu\lambda}\dot x^\mu\dot x^\nu+g_{\mu\lambda}\ddot x^\mu=\frac{1}{2}\partial_\nu g_{\mu\lambda}\dot x^\mu\dot x^\nu+\frac{1}{2}\partial_\mu g_{\nu\lambda}\dot x^\mu\dot x^\nu+g_{\mu\lambda}\ddot x^\mu$$
where in the last step we split the first term apart and rearranged indices. Putting it all together, we obtain
$$g_{\mu\lambda}\ddot x^\mu=-\frac{1}{2}\left(\partial_\nu g_{\mu\lambda}+\partial_\mu g_{\nu\lambda}-\partial_\lambda g_{\mu\nu}\right)\dot x^\mu\dot x^\nu=-\Gamma_{\lambda\mu\nu}\dot x^\mu\dot x^\nu$$
where in the last step we used the definition of the Christoffel symbols with three lower indices. Now contract with the inverse metric to raise the first index and cancel the metric on the LHS. We obtain
$$\ddot x^\lambda=-\Gamma^\lambda{}_{\mu\nu}\dot x^\mu\dot x^\nu$$
as was to be shown.
I'll try to give a slightly more precise view which might be confusing at first but will pay off later.
You are minimizing the action $S[\psi,\psi^*]$ where the square brackets indicate that $S$ is a functional: an object which takes a function as argument and spits out a number. Here $\psi$ and $\psi^*$ are still considered completely independent functions. In the following equations I will drop the dependence on $\psi^*$ since otherwise it would be too long but you can just as easily put them back in. The action then looks like
$$S[\psi]=\int\mathrm dx\,\mathcal L(\psi,\dot\psi,\nabla\psi,\nabla\dot\psi)\tag{1}$$
where $x$ is four vector so it includes time as well. Here I emphasize that $\mathcal L$ is just a regular function: it takes in 4 arguments and spits out a number. I could also fill in $a,b,c,d$ instead of $\psi,\dot\psi,\nabla\psi,\nabla\dot\psi$ in the Lagrangian (I will use $\psi$ instead of $\psi^*$ for brevity).
$$\mathcal L(a,b,c,d)=\frac i 2ab-\frac 1{2m}dc\tag{2}$$
Now to minimize the action we will have to calculate the functional derivative. The functional derivative, $\frac{\delta S}{\delta\psi}( x)$, is formally defined as
$$\int\mathrm d x\frac{\delta S}{\delta \psi}( x)\eta(x)=\lim_{\epsilon\rightarrow\infty}\frac{S[\psi+\epsilon\eta]-S[\psi]}{\epsilon}\tag{3}$$
where $\eta( x)$ is an arbitrary test function that satisfies the right boundary conditions, in this case the condition that $\eta$ goes to zero as $ x\rightarrow\infty$. $\eta$ is often written as $\delta\psi$. We can now calculate $S[\psi+\epsilon\eta]$ by Taylor expanding the Lagrangian:
\begin{align}
S[\psi+\epsilon\eta]=&\int\mathrm d x\,\mathcal L(\psi+\epsilon\eta,\dot\psi+\epsilon\dot\eta,\nabla\psi+\epsilon\nabla\eta,\nabla\dot\psi+\epsilon\nabla\dot\eta)\\
&=\int\mathrm dx\left[\mathcal L(\psi,\dot\psi,\nabla\psi,\nabla\dot\psi)+\\
\epsilon\,\eta(x)\frac{\partial}{\partial\psi}\mathcal L(\psi,\dot\psi,\nabla\psi,\nabla\dot\psi)+\\
\epsilon\,\dot\eta(x)\frac{\partial}{\partial\dot\psi}\mathcal L(\psi,\dot\psi,\nabla\psi,\nabla\dot\psi)+\\
\epsilon\nabla\eta(x)\frac{\partial}{\partial\nabla\psi}\mathcal L(\psi,\dot\psi,\nabla\psi,\nabla\dot\psi)+\\
\epsilon\nabla\dot\eta(x)\frac{\partial}{\partial\nabla\dot\psi}\mathcal L(\psi,\dot\psi,\nabla\psi,\nabla\dot\psi)
\right]\tag{4}
\end{align}
This almost has the form of the LHS of (2) but $\eta$ still has derivatives acting on it. We can fix this by integrating by parts to move the derivative from $\eta$ to the other factor at the cost of introducing a minus sign. Plugging this into (3) gives
\begin{align}\int\mathrm d x\frac{\delta S}{\delta \psi}( x)\eta(\mathbf x)&=\int\mathrm dx\left[\frac{\partial\mathcal L}{\partial\psi}-\frac{\partial}{\partial t}\frac{\partial\mathcal L}{\partial\dot\psi}-\nabla\cdot\left(\frac{\partial\mathcal L}{\partial\nabla\psi}\right)\pm\frac{\partial}{\partial t}\nabla\cdot\left(\frac{\partial\mathcal L}{\partial\nabla\dot\psi}\right)\right]\eta(x)
\end{align}
We recognize the term between brackets as the functional derivative and setting it to zero will give us the EL equations. I'm not so sure if the $\pm$ should be plus or minus but I think it's $+$. If we had included $\psi^*$ as a separate field from the start we would have got a similar equation but with $\psi^*$ instead of $\psi$.
I believe there's an error in your Lagrangian and that $\nabla\dot\psi$ should be $\nabla\psi$. The factor $\frac i 2\psi^*\dot\psi$ should either be $i\psi^*\dot\psi$ or $\frac i 2(\psi^*\dot\psi-\psi\dot\psi^*)$. See this question or this question. If you now calculate the EL you get the desired formula.
Best Answer
As user JamalS correctly points out in his answer:
Quantum actions in QFT are allowed to have $\hbar$-dependence.
If we merely want a stationary action principle for the TDSE, and view the action functional as just a mathematical tool without physical consequences beyond the EL equations, then the $\hbar$-dependence doesn't matter.
However, perhaps OP's discomfort with Mahan's TDSE derivation is spurred by the following deeper question:
That's a good question. The answer is that there are implicit hidden $\hbar$-dependence, i.e. one should rescale the variables $$\psi~=~\frac{\psi_2}{\sqrt{\hbar}},\qquad m~=~\hbar m_2,\qquad U~=~\hbar U_2,\tag{2}$$ to obtain a classical ($\hbar$-independent) action $$\begin{align} S~=~&\int \! \mathrm{d}t ~\mathrm{d}^3r \left( i\hbar \psi^{\ast}\dot{\psi}-\frac{\hbar^2}{2m} |\nabla\psi|^2 -U|\psi|^2 \right)\cr ~\stackrel{(2)}{=}~&\int \! \mathrm{d}t ~\mathrm{d}^3r \left( i \psi_2^{\ast}\dot{\psi}_2-\frac{1}{2m_2} |\nabla\psi_2|^2 -U_2|\psi_2|^2 \right) ,\end{align}\tag{3}$$ and to restore a correction loop expansion.
--
$^1$ For the semiclassical limit, see e.g. this Phys.SE post.
$^2$ For the $\hbar$/loop-expansion, see e.g. this Phys.SE post.
$^3$ Here the subscript 2 refers to a properly normalized second-quantized formulation.