= Solution
Interpret derivatives on the space of symmetric <strains>, with inner product $A:B=A_{ij}B_{ij}$. Assume a differentiable <strain energy density> with local <stress> $\sigma=D_\varepsilon W$, and a differentiable stable equilibrium branch under the prescribed affine boundary <displacement>. Thus $u_E(x)=Ex+v_E(x)$ with $v_E=0$ on $\partial\Omega$; part(a) ensures that the mean <strain> is exactly $E$.
Vary $E$ in any symmetric direction $H$. The induced displacement variation has the form $\dot u=Hx+\dot v$, where $\dot v$ vanishes on the boundary. Differentiating the <effective elastic energy> and using <static elastic equilibrium> gives
$$
\begin{aligned}
DW^{\mathrm{eff}}(E):H
&=\frac1{|\Omega|}\int_\Omega\sigma:\operatorname{sym}\nabla\dot u\,dx\\
&=\langle\sigma\rangle:H+
\frac1{|\Omega|}\int_\Omega\sigma:\operatorname{sym}\nabla\dot v\,dx\\
&=\langle\sigma\rangle:H.
\end{aligned}
$$
The last term is zero by <integration by parts>, since the <body force> is zero and $\dot v$ has zero boundary trace. As $H$ is arbitrary,
$$
\boxed{DW^{\mathrm{eff}}(E)=\langle\sigma\rangle.}
$$
This is a work-conjugacy statement on symmetric <tensors>; treating off-diagonal entries as independent scalar coordinates without their contraction weights would introduce erroneous factors of two. If the minimized <effective elastic energy> has a nondifferentiable point, the corresponding formulation uses a <subgradient> rather than a claimed classical derivative.
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