Question: How do you solve the equation $\frac{d\Phi}{dr} = \frac{m+4\pi r^{3} \rho}{r(r-2m)}$ for $\Phi$ as a function of $r$?
Note:
Here $m, \rho$ just represents the mass and density which can be taken as some constant value.
If it helps the answer is $\Phi(r) = \frac{1}{2} \log\left({1-\frac{2m}{r}}\right) + \Phi_0 $ where I have used $\Phi_0$ as the constant of integration.
Your solution is correct just for $\rho=0$. Anyway, this equation is straightforwardly integrable when $\rho$ is a constant otherwise you have to write the general solution in integral form $$ \Phi(r)=\int \frac{m+4\pi r^{3} \rho(r)}{r(r-2m)}dr=\frac{1}{2}\ln\left(1-\frac{2m}{r}\right)+4\pi\int\frac{r^2\rho(r)}{r-2m}+\Phi_0. $$ Finally, I will give you the solution for $\rho$ constant that is an elementary integral $$ \Phi(r)=\frac{1}{2}\ln\left(1-\frac{2m}{r}\right)+2\pi r^2\rho+8\pi mr\rho+16\pi m^2\rho\ln\left(\frac{r}{2m}-1\right)+\Phi_0. $$ Everything is properly defined provided $r>2m$.