Answer
$x= e$ is a local maximum
Work Step by Step
Given $$ g(x)=\frac{\ln x}{x}$$
Since
\begin{align*}
g^{\prime}(x)&=\frac{\frac{d}{dx}\left(\ln \left(x\right)\right)x-\frac{d}{dx}\left(x\right)\ln \left(x\right)}{x^2} \\
&= \frac{x(1 / x)-\ln x}{x^{2}}\\
&=\frac{1-\ln x}{x^{2}}
\end{align*}
Then $g(x)$ has critical points when
\begin{align*}
g'(x)&=0\\
\frac{1-\ln x}{x^{2}}&= 0\\
1-\ln x&=0
\end{align*}
Then $x= e$ is a critical point. Now, we use the second derivative to check $x= e$
\begin{align*}
g''(x) &= \frac{\frac{d}{dx}\left(1-\ln \left(x\right)\right)x^2-\frac{d}{dx}\left(x^2\right)\left(1-\ln \left(x\right)\right)}{\left(x^2\right)^2}\\
&= -\frac{-2\ln \left(x\right)+3}{x^3}
\end{align*}
Hence $$g''(e)= -\frac{-2\ln \left(e\right)+3}{e^3}= \frac{-1}{e^3}<0 $$
Since the second derivative at the critical point is negative, then $x= e$ is a local maximum.
