= Solution
Assume $r_0$ is an interior point and choose $\varepsilon$ small enough that its closed ball lies inside $V$. Apply <Green's second identity> to the punctured region $V_\varepsilon=V\setminus\overline{B_\varepsilon(r_0)}$, where the potential $\psi=1/|r-r_0|$ is smooth and satisfies <Laplace equation>. Since $\nabla^2\phi=-\rho$, this gives
$$
\int_{V_\varepsilon}\frac{\rho(r)}{|r-r_0|}\,dV
=\int_S(\phi\partial_n\psi-\psi\partial_n\phi)\,dS
+\int_{S_\varepsilon}(\phi\partial_{n_\varepsilon}\psi-\psi\partial_{n_\varepsilon}\phi)\,dS.
$$
The crucial orientation is the outward unit <normal vector> of the punctured region: on the inner <sphere>, it is $n_\varepsilon=-e_r$, pointing toward $r_0$. Consequently $\psi=1/\varepsilon$ and $\partial_{n_\varepsilon}\psi=1/\varepsilon^2$. The inner boundary term becomes
$$
\int_{S_\varepsilon}\left(\frac{\phi}{\varepsilon^2}+\frac1\varepsilon\partial_r\phi\right)dS.
$$
Writing $r=r_0+\varepsilon\omega$ and $dS=\varepsilon^2d\Omega$, its first part tends to $\int_{|\omega|=1}\phi(r_0)\,d\Omega=4\pi\phi(r_0)$ by continuity. The second part is $O(\varepsilon)$ because the <gradient> of $\phi$ is bounded near $r_0$, and hence tends to zero.
The omitted volume term also tends to zero: for bounded $\rho$ near $r_0$, its absolute value is bounded by a constant times $4\pi\int_0^\varepsilon r\,dr=O(\varepsilon^2)$. Thus the limiting <integral> exists, and rearranging the outer boundary terms gives
$$
\boxed{4\pi\phi(r_0)
=\int_V\frac{\rho(r)}{|r-r_0|}\,dV
+\int_S\left[\frac1{|r-r_0|}\partial_n\phi
-\phi\partial_n\frac1{|r-r_0|}\right]dS.}
$$
This is <Green's third identity> for the <Poisson equation> with the stated sign convention. It retains both boundary terms; no homogeneous boundary condition has been assumed. The punctured-domain calculation accounts for the singular source without incorrectly applying the regular <divergence theorem> through $r_0$.
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