= Solution
To first order, invert $m_0^2=m^2[1+g/(8\pi^2\epsilon)]$ and use $g=\lambda_0\mu^{-\epsilon}+O(\lambda_0^2)$:
$$
\boxed{m^2=m_0^2\left[1-\frac{\lambda_0\mu^{-\epsilon}}{8\pi^2\epsilon}\right]+O(\lambda_0^2).}
$$
The remainder is a perturbative statement at fixed regulator. When differentiating, hold the <bare mass> $m_0^2$ fixed as well as $\lambda_0$; otherwise the <mass> running would not be defined by a fixed bare theory. Since $\mu\,\partial_\mu\mu^{-\epsilon}=-\epsilon\mu^{-\epsilon}$,
$$
\mu\frac{d\log m^2}{d\mu}\bigg|_{\mathrm{bare}}
=\frac{\lambda_0\mu^{-\epsilon}}{8\pi^2}+O(g^2)
=\frac{g}{8\pi^2}+O(g^2).
$$
Thus, with the source's extra factor of one half,
$$
\boxed{\gamma_m(g)=\frac{g}{16\pi^2}+O(g^2).}
$$
The factor $-\epsilon$ from the running dimensionless <bare coupling> cancels the <counterterm> pole. Differentiating as though $g$ were independent of $\mu$ would miss this finite result. Some conventions call $\mu\,d\log m^2/d\mu$ itself the <mass> <anomalous dimension>; that convention is twice the one defined here.
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