Answer is: option4

Solution:

Using the quotient rule and chain rule:

$$f^{\prime }(x)=\frac{d}{dx}\left(\frac{(\ln x{)}^2}{x}\right)=\frac{2\ln x\cdot \frac{1}{x}\cdot x-(\ln x{)}^2\cdot 1}{x^2}=\frac{2\ln x-(\ln x{)}^2}{x^2}$$

Simplify:

$$f^{\prime }(x)=\frac{\ln x(2-\ln x)}{x^2}$$

Find critical points by setting $f^{\prime }(x)=0$:

$$\frac{\ln x(2-\ln x)}{x^2}=0$$

Since $x^2\ne 0$ for $x>0$, we solve:

$$\ln x(2-\ln x)=0$$

This gives two solutions:

1. $\ln x=0\Rightarrow x=1$
2. $2-\ln x=0\Rightarrow \ln x=2\Rightarrow x=e^2$

Determine the nature of the critical points using the second derivative or first derivative test:

For $x=1$:

1. Test $x$ slightly less than 1 (e.g., $x=0.5$):
   $\ln 0.5<0$ and $2-\ln 0.5>0$, so $f^{\prime }(0.5)<0$
2. Test $x$ slightly greater than 1 (e.g., $x=2$):
   $\ln 2>0$ and $2-\ln 2>0$, so $f^{\prime }(2)>0$

Since $f^{\prime }(x)$ changes from negative to positive, $x=1$ is a relative minimum.

For $x=e^2$:

1. Test $x$ slightly less than $e^2$ (e.g., $x=e$):
   $\ln e=1>0$ and $2-\ln e=1>0$, so $f^{\prime }(e)>0$
2. Test $x$ slightly greater than $e^2$ (e.g., $x=e^3$):
   $\ln e^3=3>0$ but $2-\ln e^3=-1<0$, so $f^{\prime }(e^3)<0$

Since $f^{\prime }(x)$ changes from positive to negative, $x=e^2$ is a relative maximum.

Conclusion: The relative maximum occurs at $x=e^2$, which corresponds to option (D).