= Solution
Contract the <tensors>, counting both off-diagonal shear terms:
$$
\mathcal W=\frac12\tau_{ij}e_{ij}
=\frac{Ez^2}{2(1-\nu^2)}
\{a^2+2\nu ac+c^2+2(1-\nu)b^2\}.
$$
Since $\int_{-h/2}^{h/2}z^2\,dz=h^3/12$, integration gives
$$
\boxed{\int_{-h/2}^{h/2}\mathcal W\,dz
=\frac D2\{a^2+2\nu ac+c^2+2(1-\nu)b^2\},\qquad
D=\frac{Eh^3}{12(1-\nu^2)}.}
$$
For the general midplane deflection, expand the energy <density> as $w_{xx}^2+2\nu w_{xx}w_{yy}+w_{yy}^2+2(1-\nu)w_{xy}^2$. A variation $w+\varepsilon\eta$ gives
$$
\delta\mathcal U=D\iint_\Omega
\{(w_{xx}+\nu w_{yy})\eta_{xx}
+(w_{yy}+\nu w_{xx})\eta_{yy}+2(1-\nu)w_{xy}\eta_{xy}\}\,dx\,dy.
$$
For clamped variations, $\eta=\partial_n\eta=0$ on the boundary, so integrating twice transfers all derivatives to $w$ without a boundary contribution. The bulk coefficient is
$$
w_{xxxx}+2\nu w_{xxyy}+w_{yyyy}+2(1-\nu)w_{xxyy}
=(\partial_x^2+\partial_y^2)^2w.
$$
Stationarity for all compactly supported variations and the fundamental lemma therefore give
$$
\boxed{\nabla^4w=0\quad\text{in the plate interior}.}
$$
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