Answer
\[\text{Relative maximum at }x=\frac{1}{3}\sqrt{e}\]
Work Step by Step
\[\begin{align}
& f\left( x \right)=\frac{\ln \left( 3x \right)}{2{{x}^{2}}} \\
& \text{Calculating the first derivative} \\
& f'\left( x \right)=\frac{d}{dx}\left[ \frac{\ln \left( 3x \right)}{2{{x}^{2}}} \right] \\
& f'\left( x \right)=\frac{2{{x}^{2}}\frac{d}{dx}\left[ \ln \left( 3x \right) \right]-\ln \left( 3x \right)\frac{d}{dx}\left[ 2{{x}^{2}} \right]}{{{\left( 2{{x}^{2}} \right)}^{2}}} \\
& f'\left( x \right)=\frac{2{{x}^{2}}\left( \frac{1}{x} \right)-\ln \left( 3x \right)\left( 4x \right)}{4{{x}^{4}}} \\
& f'\left( x \right)=\frac{2x-4x\ln \left( 3x \right)}{4{{x}^{4}}} \\
& f'\left( x \right)=\frac{1-2\ln \left( 3x \right)}{2{{x}^{3}}} \\
& \text{Calculating the critical points, set }f'\left( x \right)=0 \\
& f'\left( x \right)=\frac{1-2\ln \left( 3x \right)}{2{{x}^{3}}} \\
&1-2\ln \left( 3x \right)=0 \\
& \ln \left( 3x \right)=\frac{1}{2} \\
& 3x={{e}^{1/2}} \\
& x=\frac{1}{3}\sqrt{e} \\
& \text{The domain of the function is }\left( 0,\infty \right),\text{ then} \\
& \text{Set the intervals }\left( 0,\frac{1}{3}\sqrt{e} \right),\left( \frac{1}{3}\sqrt{e},\infty \right) \\
& \text{Making a table of values} \\
& \begin{matrix}
\text{Interval} & \left( 0,\frac{1}{3}\sqrt{e} \right) & \left( \frac{1}{3}\sqrt{e},\infty \right) \\
\text{Test Value} & x=\frac{1}{2} & x=2 \\
\text{Sign of }f'\left( x \right) & \text{ }f'\left( \frac{1}{2} \right)>0 & \text{ }f'\left( 2 \right)<0 \\
\text{Conclusion} & \text{Increasing} & \text{Decreasing} \\
\end{matrix} \\
& \text{Therefore,} \\
& \text{Relative maximum at }x=\frac{1}{3}\sqrt{e} \\
\end{align}\]
