If $\int\left\{\log (\log x)+\frac{1}{(\log x)^2}\right\} d x=x\{f(x)-g(x)\}+C$, then

$f(x)=\log (\log x) ; g(x)=\frac{1}{\log x}$

We have,

$I =\int\left\{\log (\log x)+\frac{1}{(\log x)^2}\right\} d x$

$\Rightarrow I =\int e^t\left(\log t+\frac{1}{t^2}\right) d t, \text { where } t=\log x$

$\Rightarrow I =\int e^t\left(\log t+\frac{1}{t}\right) d t+\int e^t\left(-\frac{1}{t}+\frac{1}{t^2}\right) d t$

$\Rightarrow I =e^t \log t+e^t\left(-\frac{1}{t}\right)+C$

$\Rightarrow I=x\left(\log (\log x)-\frac{1}{\log x}\right)+C$

∴  $f(x)=\log(\log (x))$ and $g(x)=\frac{1}{\log x}$