= Solution
The derivatives needed for varying $A_\nu$ are
$$
\frac{\partial\hat{\mathcal L}}{\partial(\partial_\mu A_\nu)}=-F^{\mu\nu},
\qquad
\frac{\partial\hat{\mathcal L}}{\partial A_\nu}
=-iq(\phi^*D^\nu\phi-\phi\bar D^\nu\phi^*)=-J^\nu.
$$
Therefore the electromagnetic <Euler-Lagrange equation> is
$$
\boxed{\partial_\mu F^{\mu\nu}=J^\nu,\qquad
J^\mu=iq(\phi^*D^\mu\phi-\phi\bar D^\mu\phi^*).}
$$
Antisymmetry of $F$ and commutation of derivatives imply $\partial_\nu\partial_\mu F^{\mu\nu}=0$, hence $\partial_\nu J^\nu=0$. In ordinary derivatives, $J^\mu=j^\mu-2q^2A^\mu|\phi|^2$; the interaction contribution must not be omitted.
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