Answer
$-\dfrac{1+\ln x}{(x \ln x)^2}$
Work Step by Step
We have: $f(x)=\dfrac{1}{x \ln x}=(x \ln x)^{-1}$
We differentiate both sides with respect to $x$.
$f^{\prime}(x)=\dfrac{d}{dx} [(x \ln x)^{-1}]$
Use rule: $\displaystyle \frac{d}{dx}[u^{n}]=nu^{n-1} \frac{du}{dx}$
Let us consider that $a=\ln |x|$
Now, $f^{\prime}(x)=- (x \ln x)^{-2}(x \cdot \dfrac{1}{x}+\ln (x) (1)]$
Simplify to obtain:
$f^{\prime}(x)=-\dfrac{1+\ln x}{(x \ln x)^2}$
