= Solution
The inverse <Lorentz transformation of electromagnetic fields>, from the primed rest frame to the unprimed frame, gives
$$
\boxed{
E_x=E_x',\qquad E_y=\gamma E_y',\qquad E_z=\gamma E_z'
}
$$
and
$$
\boxed{
B_x=0,\qquad
B_y=-\frac{\gamma v}{c^2}E_z',\qquad
B_z=\frac{\gamma v}{c^2}E_y'
}.
$$
Equivalently, $\mathbf B=\gamma\,\mathbf v\times\mathbf E'/c^2$ for $\mathbf v=v\mathbf e_x$.
Using $1/\mu_0=\epsilon_0c^2$ and $\gamma^2=(1-v^2/c^2)^{-1}$,
$$
\begin{aligned}
w
&=\frac{\epsilon_0}{2}\left[
E_x'^2+\gamma^2(E_y'^2+E_z'^2)
+\gamma^2\frac{v^2}{c^2}(E_y'^2+E_z'^2)
\right]\\
&=\boxed{\frac{\epsilon_0}{2}\left[
E_x'^2+\frac{c^2+v^2}{c^2-v^2}
(E_y'^2+E_z'^2)\right]}.
\end{aligned}
$$
The primed field is static, so $w$ depends on $t,x$ through
$$
x'=\gamma(x-vt)
$$
and has no explicit $t'$ dependence. The <chain rule> gives
$$
\partial_tw=-\gamma v\,\partial_{x'}w,
\qquad
\partial_xw=\gamma\,\partial_{x'}w.
$$
Consequently
$$
\boxed{
\frac{\partial w}{\partial t}
+\nabla\mathbin{\cdot}(wv\mathbf e_x)
=\partial_tw+v\partial_xw=0
}.
$$
\b[Thus the field-energy profile is transported rigidly with velocity $v\mathbf e_x$.]
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