= Solution
For the <canonical link function>, $\theta_i=\eta_i=x_i^T\beta$. Differentiating the log-likelihood gives the <score function>
$$
\nabla_\beta\ell(\beta)=\sigma^{-2}X^T(y-\mu).
$$
Since
$$
\frac{d\mu_i}{d\eta_i}=b''(\theta_i)=V(\mu_i),
$$
where $V$ is the <variance function>, a second differentiation gives
$$
\nabla_\beta^2\ell(\beta)
=-\sigma^{-2}X^T
\operatorname{diag}\bigl(V(\mu_1),\ldots,V(\mu_n)\bigr)X.
$$
This Hessian does not depend on the observed $y$, so taking its negative expectation changes nothing. Hence
$$
\boxed{\mathcal I(\beta)=\sigma^{-2}X^TWX,
\qquad W=\operatorname{diag}\bigl(V(\mu_1),\ldots,V(\mu_n)\bigr).}
$$
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