= Solution
Let $Y_1,\ldots,Y_n$ be independent. In the general exponential-dispersion formulation, with known weights $w_i>0$ and common dispersion $\phi>0$, their densities have the form
$$
f_i(y_i;\theta_i,\phi)
=\exp\left\{\frac{w_i}{\phi}
[y_i\theta_i-b(\theta_i)]
+c\left(y_i,\frac{\phi}{w_i}\right)\right\}.
$$
Thus
$$
\mu_i=\mathbb E Y_i=b'(\theta_i),
\qquad
\operatorname{Var}(Y_i)=\frac{\phi}{w_i}b''(\theta_i)
=\frac{\phi}{w_i}V(\mu_i).
$$
The systematic component is $\eta_i=x_i^T\beta$, and a specified one-to-one differentiable link relates it to the mean by
$$
g(\mu_i)=\eta_i.
$$
The unknown model parameters are the regression coefficient $\beta$ and, when it is not fixed by the family, the dispersion $\phi$.
The log-likelihood is
$$
\ell(\beta,\phi)
=\sum_{i=1}^n\left\{
\frac{w_i}{\phi}[y_i\theta_i-b(\theta_i)]
+c\left(y_i,\frac{\phi}{w_i}\right)
\right\},
$$
up to terms independent of the parameters, where $\theta_i$ is determined from $g^{-1}(x_i^T\beta)=b'(\theta_i)$.
For a Poisson response,
$$
f(y;\mu)=\exp\{y\log\mu-\mu-\log(y!)\},
$$
so the natural parameter is $\theta=\log\mu$ and $b(\theta)=e^\theta$. The canonical link sets the linear predictor equal to the natural parameter and is therefore
$$
\boxed{g(\mu)=\log\mu}.
$$
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