Given : $f(x)=\log (\log x)+(\log x)^{-2}$
Anti-derivative of
$f(x)=\int\left(\log (\log x)+(\log x)^{-2}\right) d x$ ...(i)
Let $t=\log x \Rightarrow x=e^t$
$\Rightarrow d x=e^t d t$
Using above values in eqn. (i), we get :
Anti-derivative of $f(x)=\int e^t\left(\log t+t^{-2}\right) d t$
$\begin{aligned} & \Rightarrow \quad g(x)=\int e^t\left(\log t+t^{-1}-t^{-1}+t^{-2}\right) d t \\ & \Rightarrow \quad g(x)=\int e^t\left(\log t+t^{-1}\right) d t+\int e^t\left(-t^{-1}+t^{-2}\right) d t \\ & =e^t \log t-e^t \cdot t^{-1}+C \\ & =e^t\left(\log t-t^{-1}\right)+C\end{aligned}$
$\Rightarrow \quad g(x)=e^{(\log x)}\left(\log (\log x)-(\log x)^{-1}\right)+C$ ...(i)
$\because \quad$ Equation (i) passes through $(e, 2023-e)$
$\therefore \quad$ Eqn. (i) reduces to :
$2023-e=e^{(\log e)}\left(\log (\log e)-(\log e)^{-1}\right)+C$
$\begin{aligned} & \Rightarrow \quad 2023-e=e(\log (1)-1)+C \\ & \Rightarrow \quad 2023-e=-e+C \Rightarrow C=2023\end{aligned}$
Putting the value of $C$ in eqn. (i) :
$g(x)=e^{(\log x)}\left(\log (\log x)-(\log x)^{-1}\right)+2023$
$\therefore \quad$ Sum of digits of independent term $=2+0+2+3=7$.