Answer
$$y' = \frac{1}{{x\left( {\ln x} \right)\ln \left( {\ln x} \right)}}$$
Work Step by Step
$$\eqalign{
& y = \ln \left( {\ln \left( {\ln x} \right)} \right) \cr
& {\text{differentiate}} \cr
& y' = \left[ {\ln \left( {\ln \left( {\ln x} \right)} \right)} \right]' \cr
& {\text{chain rule}}{\text{, recall that }}\left( {\ln u} \right)' = \frac{{u'}}{u} \cr
& y' = \frac{1}{{\ln \left( {\ln x} \right)}}\left( {\ln \left( {\ln x} \right)} \right)' \cr
& y' = \frac{1}{{\ln \left( {\ln x} \right)}}\frac{1}{{\ln x}}\left( {\ln x} \right)' \cr
& y' = \frac{1}{{\ln \left( {\ln x} \right)}}\frac{1}{{\ln x}}\left( {\frac{1}{x}} \right) \cr
& {\text{simplify}} \cr
& y' = \frac{1}{{x\left( {\ln x} \right)\ln \left( {\ln x} \right)}} \cr} $$
