the equation of continuity in tensor form

$$\rho = \Sigma_a e_a \delta(r-r_a)$$ or
$$de = \rho dV$$
$$dx^i de = \rho dx^idV$$
$$dx^i de = \rho \frac{dx^i}{dt}dV dt$$
$$dx^i de = \rho v^i\frac{d\Omega}{c}$$
$$dx^i de = j^i \frac{d\Omega}{c}$$
where $j^i=(\rho c, \rho v^i)$
now, consider the term $\int \frac{e}{c}A_kdx^k$ from the action term
$$\int \frac{e}{c}A_kdx^k = \int \frac{1}{c}\rho dV A_k dx^k$$
$$= \int \frac{1}{c}\rho A_k \frac{dx^k}{dt} dV dt$$
$$= \int \frac{1}{c^2}j^k A_k d\Omega$$
substituting this in the action equation
$$S = -\int_a^b mcds - \int_a^b\frac{e}{c}A_{\mu} dx^{\mu} - \frac{1}{16\pi c}\int_a^b d\Omega F_{\mu\nu}F^{\mu\nu}$$
$$S = -\int mcds - \int \frac{1}{c^2}j^k A_k d\Omega - \frac{1}{16\pi c}\int d\Omega F_{\mu\nu}F^{\mu\nu}$$
$$\delta S = -\int mc dv^{\mu}\delta x_{\mu} - \int \frac{1}{c^2}\delta j^k A_k d\Omega - \int \frac{1}{c^2}j^k \delta A_k d\Omega -\frac{1}{16\pi c}\int d\Omega \delta (F_{\mu\nu}F^{\mu\nu})$$
$$\delta (F_{\mu\nu}F^{\mu\nu}) = 2 F_{\mu\nu}\delta F^{\mu\nu}$$
$$=2 F_{\mu\nu}\delta (d^{\mu}A^{\nu}-d^{\nu}A^{\mu})$$
$$=2 F_{\mu\nu} (2 \delta d^{\mu}A^{\nu})$$
$$=4 F_{\mu\nu} d^{\mu}\delta A^{\nu}$$
$$=4 d^{\mu}(F_{\mu\nu} \delta A^{\nu}) -4 d^{\mu}F_{\mu\nu} \delta A^{\nu}$$
$$\delta S = -\int mc dv^{\mu}\delta x_{\mu} - \int \frac{1}{c^2}\delta j^k A_k d\Omega - \int \frac{1}{c^2}j^k \delta A_k d\Omega -\frac{1}{16\pi c}\int 4( d^{\mu}(F_{\mu\nu} \delta A^{\nu}) -d^{\mu}F_{\mu\nu} \delta A^{\nu})d\Omega$$
considering the terms with a variation in $A_k$
$$\delta S = - \int \frac{1}{c^2}j^k \delta A_k d\Omega +\frac{1}{4\pi c}\int d^{\mu}F_{\mu\nu} \delta A^{\nu}d\Omega$$
since $\delta S =0$
$$- \int \frac{1}{c^2}j^k \delta A_k d\Omega +\frac{1}{4\pi c}\int d^{\mu}F_{\mu\nu} \delta A^{\nu}d\Omega = 0$$
$$- \int \frac{4\pi}{c}j^k \delta A_k d\Omega +\int d^{\mu}F_{\mu\nu} \delta A^{\nu}d\Omega = 0$$
$$\int (-\frac{4\pi}{c}j^{\nu} \delta A_{\nu} d\Omega + d^{\mu}F_{\mu\nu} \delta A^{\nu}d\Omega) = 0$$
$$\int \delta A^{\nu}d\Omega (-\frac{4\pi}{c}j^{\nu} + d^{\mu}F_{\mu\nu}) = 0$$
$$-\frac{4\pi}{c}j^{\nu} + d^{\nu}F_{\nu\mu} = 0$$
$$d^{\nu}F_{\nu\mu} = \frac{4\pi}{c}j^{\nu}$$
$$d_{\nu}F_{\mu\nu} = -\frac{4\pi}{c}j_{\nu}$$