Answer
$$ g'(x)= \frac{1}{x(1+(\ln x)^2)}.$$
Work Step by Step
Recall that $(\tan^{-1} x)'=\dfrac{1}{1+x^2}$
Recall that $(\ln x)'=\dfrac{1}{x}$
Since $ g(x)=\tan^{-1}\ln x $, then the derivative, using the chain rule, is given by
$$ g'(x)= \frac{1}{1+(\ln x)^2}(\ln x)'= \frac{1/x}{1+(\ln x)^2}= \frac{1}{x(1+(\ln x)^2)}.$$
