Answer
$$y' = {\left( {\ln x} \right)^{\ln x}}\left( {\frac{1}{x} + \frac{{\ln \left( {\ln x} \right)}}{x}} \right)$$
Work Step by Step
$$\eqalign{
& y = {\left( {\ln x} \right)^{\ln x}} \cr
& {\text{take logarithm natural }} \cr
& \ln y = \ln {\left( {\ln x} \right)^{\ln x}} \cr
& {\text{logarithm properties}} \cr
& \ln y = \ln x\ln \left( {\ln x} \right) \cr
& {\text{differentiate}} \cr
& \left( {\ln y} \right)' = \left( {\ln x\ln \left( {\ln x} \right)} \right)' \cr
& {\text{product rule}} \cr
& \frac{{y'}}{y} = \ln x\left( {\ln \left( {\ln x} \right)} \right)' + \ln \left( {\ln x} \right)\left( {\ln x} \right)' \cr
& \frac{{y'}}{y} = \ln x\left( {\frac{{\frac{1}{x}}}{{\ln x}}} \right) + \ln \left( {\ln x} \right)\left( {\frac{1}{x}} \right) \cr
& \frac{{y'}}{y} = \frac{1}{x} + \frac{{\ln \left( {\ln x} \right)}}{x} \cr
& {\text{solve for }}y' \cr
& y' = y\left( {\frac{1}{x} + \frac{{\ln \left( {\ln x} \right)}}{x}} \right) \cr
& {\text{replace }}y = {\left( {\ln x} \right)^{\ln x}} \cr
& y' = {\left( {\ln x} \right)^{\ln x}}\left( {\frac{1}{x} + \frac{{\ln \left( {\ln x} \right)}}{x}} \right) \cr} $$