Answer
$$r'\left( x \right) = \frac{2}{x} - \frac{{2\ln \left( {x - 1} \right)}}{{x - 1}}$$
Work Step by Step
$$\eqalign{
& r\left( x \right) = \ln \left( {{x^2}} \right) - {\left[ {\ln \left( {x - 1} \right)} \right]^2} \cr
& {\text{Differentiate}} \cr
& r'\left( x \right) = \frac{d}{{dx}}\left[ {\ln \left( {{x^2}} \right)} \right] - \underbrace {\frac{d}{{dx}}{{\left[ {\ln \left( {x - 1} \right)} \right]}^2}}_{{\text{Use chain rule}}} \cr
& {\text{Recall that }}\frac{d}{{dx}}\left[ {\ln u} \right] = \frac{{u'}}{u} \cr
& r'\left( x \right) = \frac{{2x}}{{{x^2}}} - 2\left[ {\ln \left( {x - 1} \right)} \right]\frac{d}{{dx}}\left[ {\ln \left( {x - 1} \right)} \right] \cr
& r'\left( x \right) = \frac{2}{x} - 2\left[ {\ln \left( {x - 1} \right)} \right]\left( {\frac{1}{{x - 1}}} \right) \cr
& {\text{Simplifying}} \cr
& r'\left( x \right) = \frac{2}{x} - \frac{{2\ln \left( {x - 1} \right)}}{{x - 1}} \cr} $$
