= Solution
Differentiate the electromagnetic <stress-energy tensor>. The current term is
$$
\frac1{\mu_0}(\partial_\mu F^{\mu\alpha})F^\nu{}_\alpha=-J^\alpha F^\nu{}_\alpha.
$$
The other two terms cancel by the homogeneous <Maxwell equations>. To see the contraction explicitly, multiply $\partial_\mu F_{\beta\alpha}+\partial_\beta F_{\alpha\mu}+\partial_\alpha F_{\mu\beta}=0$ by $F^{\mu\alpha}$. The first and third contracted terms are equal after swapping indices and using antisymmetry; the middle is $-\tfrac12\partial_\beta(F^{\mu\alpha}F_{\mu\alpha})$. Therefore
$$
F^{\mu\alpha}\partial_\mu F^\nu{}_\alpha
=\frac14\partial^\nu(F^{\alpha\beta}F_{\alpha\beta}).
$$
It cancels differentiation of the trace term, leaving
$$
\boxed{\partial_\mu T_{\rm em}^{\mu\nu}=-F^\nu{}_\mu J^\mu}.
$$
The sign is tied to the printed convention $\partial_\mu F^{\mu\nu}=-\mu_0J^\nu$ and the displayed field-strength tensor.
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