= Solution
Write $\theta_i=\theta(\mu_i)$ and $\eta_i=x_i^T\beta=g(\mu_i)$. Differentiating $\mu_i=K'(\theta_i)$ gives $d\mu_i/d\theta_i=V(\mu_i)$, so
$$
\frac{\partial\theta_i}{\partial\beta_j}
=\frac{x_{ij}}{V(\mu_i)g'(\mu_i)}.
$$
In the <independent> <exponential dispersion family>, only $[y_i\theta_i-K(\theta_i)]/\sigma_i^2$ depends on $\beta$. Its derivative therefore gives the <score function>
$$
\boxed{U_j=\sum_i\frac{(y_i-\mu_i)x_{ij}}{\sigma_i^2V(\mu_i)g'(\mu_i)}.}
$$
The residuals are <independent>, have mean zero and variance $\sigma_i^2V(\mu_i)$. Consequently their score covariance is the <Fisher information matrix>:
$$
\boxed{\mathcal I_{jk}=\mathbb E(U_jU_k)
=\sum_i\frac{x_{ij}x_{ik}}{\sigma_i^2V(\mu_i)[g'(\mu_i)]^2}.}
$$
This also equals minus the expected likelihood Hessian under the usual regularity conditions.
For the <canonical link function>, $g(\mu)=\theta(\mu)$ and $g'=1/V$, so
$$
\boxed{U_j=\sum_i\frac{(y_i-\mu_i)x_{ij}}{\sigma_i^2},\qquad
\mathcal I_{jk}=\sum_i\frac{V(\mu_i)x_{ij}x_{ik}}{\sigma_i^2}.}
$$
To obtain the <maximum-likelihood estimate>, solve $U(\widehat\beta)=0$, for example with <Fisher scoring> iterations $\beta^{new}=\beta+\mathcal I(\beta)^{-1}U(\beta)$. Equivalently use <iteratively reweighted least squares>, with weights $[\sigma_i^2V(\mu_i)g'(\mu_i)^2]^{-1}$ and working response $\eta_i+(y_i-\mu_i)g'(\mu_i)$. Since $\sigma_i^2=\sigma^2a_i$, the common dispersion scale cancels from this update for $\beta$. A nonsingular information matrix and a finite interior likelihood maximum are needed; iteration does not assert those for every possible dataset.
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