Find the intervals in which $f\left(x\right) = \dfrac{x}{\log x}$ is increasing or decreasing.
$f\left(x\right) = \dfrac{x}{\log x}$
$\therefore$ $f'\left(x\right) = \dfrac{\log x - x \times \dfrac{1}{x}}{\left(\log x\right)^2}$
i.e. $f'\left(x\right) = \dfrac{\log x -1}{\left(\log x\right)^2}$
Now, $\left(\log x\right)^2 > 0 \;\; \forall \; x $
$\therefore$ $f'\left(x\right) > 0$ $\implies$ $\log x - 1 > 0$
i.e. $\log x > 1$ $\implies$ $\log x > \log e$ $\implies$ $x>e$
$\therefore$ $f\left(x\right)$ is increasing in the interval $\left(e,\infty\right)$
$f'\left(x\right) < 0$ $\implies$ $\log x -1 < 0$
i.e. $\log x < 1$ $\implies$ $x < e$
$\therefore$ $f\left(x\right)$ is decreasing in the interval $\left(0,e\right)$