Answer
$$\frac{d}{{dx}}{\left( {\ln x} \right)^{{x^2}}} = {\left( {\ln x} \right)^{{x^2}}}\left( {\frac{x}{{\ln x}} + 2x\left( {\ln \left( {\ln x} \right)} \right)} \right)$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}{\left( {\ln x} \right)^{{x^2}}} \cr
& {\text{let }}y = {\left( {\ln x} \right)^{{x^2}}} \cr
& {\text{taking logarithm on both sides}} \cr
& \ln y = \ln {\left( {\ln x} \right)^{{x^2}}} \cr
& {\text{using logarithmic properties}} \cr
& \ln y = {x^2}\ln \left( {\ln x} \right) \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left( {\ln y} \right) = \frac{d}{{dx}}\left[ {{x^2}\ln \left( {\ln x} \right)} \right] \cr
& {\text{product rule}} \cr
& \frac{d}{{dx}}\left( {\ln y} \right) = {x^2}\frac{d}{{dx}}\left[ {\ln \left( {\ln x} \right)} \right] + \ln \left( {\ln x} \right)\frac{d}{{dx}}\left[ {{x^2}} \right] \cr
& {\text{solve derivatives}} \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = {x^2}\left( {\frac{{1/x}}{{\ln x}}} \right) + \ln \left( {\ln x} \right)\left( {2x} \right) \cr
& {\text{simplifying}} \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \frac{x}{{\ln x}} + 2x\left( {\ln \left( {\ln x} \right)} \right) \cr
& {\text{solving for }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = y\left( {\frac{x}{{\ln x}} + 2x\left( {\ln \left( {\ln x} \right)} \right)} \right) \cr
& {\text{replace }}y = {\left( {\ln x} \right)^{{x^2}}} \cr
& \frac{d}{{dx}}{\left( {\ln x} \right)^{{x^2}}} = {\left( {\ln x} \right)^{{x^2}}}\left( {\frac{x}{{\ln x}} + 2x\left( {\ln \left( {\ln x} \right)} \right)} \right) \cr} $$
