= Solution
In the same <Minkowski metric> and electromagnetic units, $T^{\mu\nu}=-F^\mu{}_{\lambda}F^{\nu\lambda}+\tfrac14g^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}$. Differentiating gives
$$
\partial_\mu T^{\mu\nu}=-(\partial_\mu F^\mu{}_{\lambda})F^{\nu\lambda}-F^\mu{}_{\lambda}\partial_\mu F^{\nu\lambda}+\frac14\partial^\nu(F_{\alpha\beta}F^{\alpha\beta}).
$$
To simplify the last two terms, contract $\partial_\mu F_{\nu\lambda}+\partial_\nu F_{\lambda\mu}+\partial_\lambda F_{\mu\nu}=0$ with $F^{\mu\lambda}$. Its first and third terms are equal, and its middle term is $-\tfrac12\partial_\nu(F_{\alpha\beta}F^{\alpha\beta})$. Raising $\nu$ therefore gives $F^\mu{}_{\lambda}\partial_\mu F^{\nu\lambda}=\tfrac14\partial^\nu(F_{\alpha\beta}F^{\alpha\beta})$. The cancellation leaves
$$
\partial_\mu T^{\mu\nu}=-F^{\nu\lambda}\partial_\mu F^\mu{}_{\lambda},\qquad \boxed{\partial_\mu T^{\mu\nu}=0\quad\text{in vacuum}.}
$$
This is the <electromagnetic stress-energy divergence identity> in rationalized units. As a sign check, $T^{00}=(E^2+B^2)/2$ is positive and $T^{0i}=(\mathbf E\times\mathbf B)_i$ is the <Poynting vector>. The integrated <four-momentum> $P^\nu=\int T^{0\nu}d^3x$ is conserved if the boundary flux vanishes. With matter <sources>, the <electromagnetic field> alone instead obeys $\partial_\mu T^{\mu\nu}=-F^{\nu\lambda}J_\lambda$: <energy> and momentum transfer to matter restore conservation of the total <stress-energy tensor>.
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