= Solution
Choose a small positively oriented circle $C_\varepsilon$ about $z_0$ inside the given region. Cauchy's theorem on the region between the two <contours> gives
$$
\int_\gamma\frac{f(z)}{z-z_0}\,dz=\int_{C_\varepsilon}\frac{f(z)}{z-z_0}\,dz=i\int_0^{2\pi}f(z_0+\varepsilon e^{it})\,dt.
$$
<Continuity> makes the last integrand converge uniformly to $f(z_0)$ as $\varepsilon\downarrow0$. Its limit is $2\pi if(z_0)$, proving the <Cauchy integral formula>.
The same formula holds with an interior point $w$ near $z_0$. Subtract its value at $z_0$ and divide by $w-z_0$:
$$
\frac{f(w)-f(z_0)}{w-z_0}=\frac1{2\pi i}\int_\gamma\frac{f(z)}{(z-w)(z-z_0)}\,dz.
$$
The <contour> is a positive distance from $z_0$, so the integrands converge uniformly as $w\to z_0$. Passing to the limit gives the differentiated formula
$$
\boxed{f'(z_0)=\frac1{2\pi i}\int_\gamma\frac{f(z)}{(z-z_0)^2}\,dz.}
$$
On the <unit circle>, $\bar z=1/z$ and $\operatorname{Im}z=(z-\bar z)/(2i)$, proving the requested identity. Hence the requested integral is
$$
\frac1{2i}\left[\int_{|z|=1}\frac{dz}{3z-1}-\int_{|z|=1}\frac{dz}{z^2(3z-1)}\right].
$$
The first integral is $2\pi i/3$. For the second use
$$
\frac1{z^2(3z-1)}=-\frac1{z^2}-\frac3z+\frac9{3z-1}.
$$
The differentiated formula applied to the constant function gives $\int z^{-2}\,dz=0$; the remaining two terms integrate to $-6\pi i+6\pi i=0$. Therefore
$$
\boxed{\int_{|z|=1}\frac{\operatorname{Im}z}{z(3z-1)}\,dz=\frac\pi3.}
$$
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