= Solution
Fix the observation point $\mathbf x$ and apply <Green's second identity> in the source variable $\mathbf y$. Since $\Delta u=0$, $\Delta_{\mathbf y}G_D=\delta_{\mathbf x}$ and $G_D=0$ on the boundary, it gives
$$
u(\mathbf x)=\int_D(u\Delta G_D-G_D\Delta u)\,d^2y
=\int_{\partial D}f(\mathbf y)\partial_{n_{\mathbf y}}G_D(\mathbf x,\mathbf y)\,ds_{\mathbf y}.
$$
The identity can be justified by deleting a small circle about $\mathbf x$ and letting its radius tend to zero; the logarithmic flux contributes $u(\mathbf x)$.
Set $\mathbf x=(r,\phi)$ and $\mathbf y=(\rho,\phi')$, and write $c=\cos(\phi-\phi')$. The symmetric expression for the <Green's function> is
$$
G_D=\frac1{4\pi}\log\frac{r^2+\rho^2-2r\rho c}{1+r^2\rho^2-2r\rho c}.
$$
At the boundary, the outward <normal derivative> is $\partial_\rho$ at $\rho=1$, and hence
$$
\left.\partial_\rho G_D\right|_{\rho=1}
=\frac1{4\pi}\frac{(2-2rc)-(2r^2-2rc)}{1+r^2-2rc}
=\frac1{2\pi}\frac{1-r^2}{1-2rc+r^2}.
$$
Since $ds=d\phi'$ on the <unit circle>, the <Poisson integral on the unit disk> follows, with <Poisson kernel>
$$
\boxed{P(r,\vartheta)=\frac{1-r^2}{1-2r\cos\vartheta+r^2}},\qquad
u(r,\phi)=\frac1{2\pi}\int_0^{2\pi}f(\phi')P(r,\phi-\phi')\,d\phi'.
$$
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