Prove that $f(x_0)>\frac{2}{3}$

Hint

Try to expand $f$ at the first order around $2$ based on

$g(x) = \frac{x}{x+1} = \frac{2}{3}(1+h/6) +o(h^2)$ where $x=2+h$ and $\Gamma(2+h)=1+(1-\gamma)h+o(h^2)$ where $\gamma$ is the Euler Mascheroni constant.

Therefore $$\begin{aligned} \ln f(2+h) &= (1+(1-\gamma)h+o(h^2))(\ln(2/3) + h/6 + o(h^2))\\ &=\ln(2/3) + ((1-\gamma)\ln(2/3) + 1/6)h +o(h^2) \end{aligned} $$ proving that $f$ takes around $2$ values larger than $2/3$ as $(1-\gamma)\ln(2/3) + 1/6 \neq 0$.