Answer
$x= e^{1/3}$ is a local maximum.
Work Step by Step
Given $$ g(x)=\frac{\ln x}{x^{3}}$$
Since
\begin{align*}
g^{\prime}(x)&= \frac{\frac{d}{dx}\left(\ln \left(x\right)\right)x^3-\frac{d}{dx}\left(x^3\right)\ln \left(x\right)}{\left(x^3\right)^2}\\
&=\frac{1-3\ln \left(x\right)}{x^4}
\end{align*}
Then $g(x)$ has critical points when
\begin{align*}
g'(x)&=0\\
\frac{1-3\ln \left(x\right)}{x^4}&= 0 \\
1-3\ln \left(x\right)&=0
\end{align*}
Then $x= e^{1/3}$ is a critical point.
Now, we use the second derivative to check $x= e^{1/3} $
\begin{align*}
g''(x) &=\frac{\frac{d}{dx}\left(1-3\ln \left(x\right)\right)x^4-\frac{d}{dx}\left(x^4\right)\left(1-3\ln \left(x\right)\right)}{\left(x^4\right)^2}\\
&= -\frac{-12\ln \left(x\right)+7}{x^5}
\end{align*}
Hence $$g''(e^{1/3})=\frac{-7+12 \ln e^{1 / 3}}{\left(e^{1 / 3}\right)^{5}}=-0.5666<0 $$
Since the second derivative at the critical point is negative, we have a maximum.
Thus, $x= e^{1/3}$ is a local maximum.
