The maximum value of $f(x)=\frac{\log x}{x}$  is

The given function is $f(x)=\frac{\log x}{x}$

Differentiate both sides with respect to $x$

$f\text{'}\left(x\right)=\frac{1-\log x}{x^2}$

By first derivative test,  the funciton has maximum value when the sign of first derivative will change from positive to negative

Hence, the function has maximum value at $x=e$

Therefore, the maximum value of the given function is $\boxed{\frac{1}{e}}$