We can try the first derivative test.
$$ f(x)=\frac{\ln x}{x} $$
$$ f'(x)= \frac{1-\ln x}{x^2} $$
\(f'(x)=0\)
$$ 0=1-\ln x $$
$$ x=e $$
\(f'(x)\) is undefined
$$ x = 0 $$
The critical values are \(e\) and \(0\). Testing these intervals:
$$ \begin{array}{|c|c|c|c|} \hline
\text{Interval} & -\infty \lt x \lt 0 & 0 \lt x \lt e & e \lt x \lt \infty \\ \hline
\text{ Test Value} & x=-1 & x=\frac{e}{2} & x= e^2 \\ \hline
\text{ Sign of }f'(x) & f'(-11) \lt 0 & f'(0) \gt 0 & f'(11) \lt 0 \\ \hline
\text{Conclusion} & \text{Decreasing} & \text{Increasing} & \text{Decreasing} \\ \hline
\end{array} $$
It would appear that \(f\) does not obtain a maximum since it is decreasing on the interval \(-\infty \lt x \lt 0\).
However, recall that the domain of \(f\) is \(x \gt 0\) since it is composed of a logrithm function.
Therefore, the graph increases until it obtains its absolute maximum at \(x=e\).
$$ f(e) = \frac{\ln (e)}{e} $$
$$ f(e)= \boxed{\frac{1}{e}} $$