Answer
$x=e$ and $y=\dfrac{1}{e^2}$ giving the maximum area $A=\frac{1}{e}$
Work Step by Step
Let the area of a rectangle be $a=xy$
Here, $a= x(\dfrac{\ln x}{x^2})=\dfrac{\ln x}{x}$
Then $a'=\dfrac{1-\ln x}{x^2}=0$
This implies that $x=e$
Now, $y=(\dfrac{\ln x}{x^2})=(\dfrac{\ln e}{e^2})=\dfrac{1}{e^2}$
Hence, $x=e$ and $y=\dfrac{1}{e^2}$ giving the maximum area $A=\frac{1}{e}$
