$\mathrm{f}\left(\mathrm{x}\right)=\left|{\mathrm{xlog}}_{\mathrm{e}}\mathrm{x}\right|$monotonically decreases in

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\left|{\mathrm{xlog}}_{\mathrm{e}}\mathrm{x}\right| \\ \mathrm{For} \mathrm{g}\left(\mathrm{x}\right)=\mathrm{x} {\log }_{\mathrm{e}}\mathrm{x}, \end{gathered}$$
${\mathrm{g}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=\mathrm{x}\frac{1}{\mathrm{x}}+{\log }_{\mathrm{e}}\mathrm{x}=1+{\log }_{\mathrm{e}}\mathrm{x}$
Thus, g (x) increases for $\left(\frac{1}{\mathrm{e}},\mathrm{\infty }\right)$and decreases for $\left(0,\frac{1}{\mathrm{e}}\right)$.
 

From the graph,$\mathrm{f}\left(\mathrm{x}\right)=\left|{\mathrm{xlog}}_{\mathrm{e}}\mathrm{x}\right|$ decreases in $\left(0, \frac{1}{\mathrm{e}}\right)$.