starting from reflection formula and digamma function
I obtain \begin{align} \frac{\zeta'(s)}{\zeta(s)} + \frac{\zeta'(1-s)}{\zeta(1-s)} = \log(2\pi) + \frac{\pi}{2} \cot\left(\frac{\pi s}{2}\right) - \psi(1-s) \end{align} taking limit goes 1, reference says that \begin{align} \frac{\zeta'(0)}{\zeta(0)} = \log(2\pi) \end{align}
It seems that $\zeta'(1)=0$ and $\psi(0)=0$. I wonder how to prove this
Edit
due to @Mindlack I got $\frac{\zeta'(1)}{\zeta(1)}=0$, but still got problem with $\psi(0)=0$....
From Mathematica I see that $\psi(0) = -\infty$....
The reference I follow is Nicolas M Robles master thesis on zeta function regularization page 28
A delicate calculation that requires justification \begin{align} \frac{\mathrm{d} \zeta}{\mathrm{d}z} &= \frac{\mathrm{d} }{\mathrm{d}z}\left( \sum_{n=1}^{\infty}n^{-z}\right) \\ &= \sum_{n=1}^{\infty}\left(- \ln(n) n^{-z} \right) \\ &= -\sum_{n=1}^{\infty}\left(\frac{\ln(n)}{n^{z}} \right) \\ &= -\sum_{n=2}^{\infty}\left(\frac{\ln(n)}{n^{z}} \right) \end{align} As $\ln 1 = 0$. Your result then follows. it is delicate but there is also a need to show that the sum converges uniformly and also that the sum of the derivatives converge uniformly.