Let $\mathrm{f}(x)=\frac{\log x}{x} \Rightarrow \mathrm{f}^{\prime}(x)=\frac{1}{x^2}-\frac{\log x}{x^2}$
For maximum or minimum value of $\mathrm{f}(x)$,
$\begin{aligned}
& \mathrm{f}^{\prime}(x)=0 \\
& \Rightarrow \frac{1-\log _e x}{x^2}=0
\end{aligned}$
$\therefore \quad \log _{\mathrm{e}} x=1$ or $x=\mathrm{e}$, which lie in $(0, \infty)$.
For $x=\mathrm{e}, \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=-\frac{1}{\mathrm{e}^3}$, which is -ve .
$\therefore \quad y$ is maximum at $x=\mathrm{e}$ and its maximum value $=\frac{\log \mathrm{e}}{\mathrm{e}}=\frac{1}{\mathrm{e}}$.