= Solution
Write
$$
L=-mc\sqrt{-\dot x^2}+qA_\nu(x)\dot x^\nu.
$$
Since
$$
\sqrt{-\dot x^2}=c\frac{d\tau}{d\lambda},
\qquad
\dot x^\mu=\frac{d\tau}{d\lambda}u^\mu,
$$
its derivatives are
$$
\frac{\partial L}{\partial\dot x^\mu}
=m u_\mu+qA_\mu,
\qquad
\frac{\partial L}{\partial x^\mu}
=q(\partial_\mu A_\nu)\dot x^\nu.
$$
The <Euler-Lagrange equation> therefore gives
$$
\begin{aligned}
0
&=\frac d{d\lambda}(m u_\mu+qA_\mu)
-q(\partial_\mu A_\nu)\dot x^\nu\\
&=m\frac{du_\mu}{d\lambda}
+q(\partial_\nu A_\mu-\partial_\mu A_\nu)\dot x^\nu.
\end{aligned}
$$
Using the <electromagnetic field tensor>
$$
F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu
$$
and dividing by $d\tau/d\lambda$ gives
$$
\boxed{
m\frac{du_\mu}{d\tau}=qF_{\mu\nu}u^\nu
}.
$$
Raising the first index yields the equivalent requested form
$$
\boxed{
m\frac{du^\mu}{d\tau}=qF^{\mu\nu}u_\nu
}.
$$
This is the <Lorentz-force equation from the worldline action>.
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