Intervals of increase/decrease and local extrema via first derivative test

By the quotient rule, $f'(x) = \dfrac{\frac{1}{x}\cdot x - \ln(x)\cdot 1}{x^2} = \dfrac{1-\ln(x)}{x^2}$. Since $x^2 > 0$ for all $x>0$, the sign of $f'(x)$ matches the sign of $1-\ln(x)$. Setting $1-\ln(x)=0$ gives $\ln(x)=1$, so $x=e$ is the only critical number (there are no points where $f'$ is undefined within the domain $x>0$).

For $0<x<e$, $\ln(x)<1$ so $1-\ln(x)>0$, meaning $f'(x)>0$ and $f$ is increasing. For $x>e$, $\ln(x)>1$ so $1-\ln(x)<0$, meaning $f'(x)<0$ and $f$ is decreasing. Since $f'$ changes from positive to negative at $x=e$, the First Derivative Test tells us $f$ has a local maximum there, with value $f(e) = \ln(e)/e = 1/e$.