= Solution
In a region with no current, the <Ampère-Maxwell equation> and <Faraday's law> are
$$
\nabla\times\mathbf B=\mu_0\epsilon_0\frac{\partial\mathbf E}{\partial t},
\qquad
\nabla\times\mathbf E=-\frac{\partial\mathbf B}{\partial t}.
$$
Therefore
$$
\begin{aligned}
\frac{\partial w}{\partial t}
&=\epsilon_0\mathbf E\mathbin{\cdot}\partial_t\mathbf E
+\frac1{\mu_0}\mathbf B\mathbin{\cdot}\partial_t\mathbf B\\
&=\frac1{\mu_0}\left[
\mathbf E\mathbin{\cdot}(\nabla\times\mathbf B)
-\mathbf B\mathbin{\cdot}(\nabla\times\mathbf E)\right]\\
&=-\frac1{\mu_0}\nabla\mathbin{\cdot}(\mathbf E\times\mathbf B),
\end{aligned}
$$
where the last line uses the <divergence and curl of a cross product>. Thus <Poynting's theorem> takes the form
$$
\boxed{\frac{\partial w}{\partial t}+\nabla\mathbin{\cdot}\mathbf S=0},
\qquad
\boxed{\mathbf S=\frac1{\mu_0}\mathbf E\times\mathbf B}.
$$
The vector $\mathbf S$ is the <Poynting vector>.
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