Too Many Logs

If you take the derivative of $f(x)$ directly, then it could get pretty ugly. Let's simplify it first using the properties of logarithms.

$\log_3\left(\left(\dfrac{x+1}{x-1}\right)^{\ln3}\right)=\ln3\times\log_3(x+1)-\ln3\times\log_3(x-1)\\=\ln3\left(\dfrac{\ln(x+1)}{\ln3}-\dfrac{\ln(x-1)}{\ln3}\right)=\ln(x+1)-\ln(x-1)$

So $f(x)=\ln(x+1)-\ln(x-1).$ Therefore, $f'(x)=\frac{1}{x+1}-\frac{1}{x-1}=-\frac{2}{x^2-1}.$ But this is the formula for the slope of the $\textit{tangent}$ line, we want the slope of the $\textit{normal}$ line. This is equal to $\frac{x^2-1}{2}.$ Plugging in $x=3,$ the normal line to $f(x)$ at the point $(3,\ln2)$ is $\frac{8}{2}=\boxed{4}$