= Solution
Introduce a <Lagrange multiplier> $\lambda$ for the normalization and vary
$$
J[u]=\int_D\bigl(|\nabla u|^2-\lambda u^2\bigr)\,dx\,dy.
$$
For a smooth variation $u+\varepsilon\eta$ with $\eta=0$ on $\partial D$, <integration by parts> gives
$$
\left.\frac d{d\varepsilon}J[u+\varepsilon\eta]\right|_{\varepsilon=0}
=2\int_D(\nabla u\mathbin{\cdot}\nabla\eta-\lambda u\eta)
=-2\int_D(\nabla^2u+\lambda u)\eta.
$$
The <fundamental lemma of the calculus of variations> therefore yields the <Euler-Lagrange equation>
$$
\boxed{\nabla^2u+\lambda u=0}.
$$
Multiplying by $u$ and integrating, while using
$$
\nabla\mathbin{\cdot}(u\nabla u)=|\nabla u|^2+u\nabla^2u,
$$
the <divergence theorem> and $u=0$ on the boundary give
$$
0=I[u]+\int_Du\nabla^2u=I[u]-\lambda\int_Du^2.
$$
The normalization is one, so the multiplier equals the stationary value:
$$
\boxed{\lambda=I[u]}.
$$
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