The equation of a curve is $y = \frac{\ln x}{x}$.

1. Differentiate the function using the quotient rule

Let $u = \ln x$ and $v = x$.
$\frac{du}{dx} = \frac{1}{x}$ and $\frac{dv}{dx} = 1$.
$\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$
$\frac{dy}{dx} = \frac{x(\frac{1}{x}) - (\ln x)(1)}{x^2}$
$\frac{dy}{dx} = \frac{1 - \ln x}{x^2}$

2. Set the derivative to zero to find the x-coordinate of the stationary point

For stationary points, $\frac{dy}{dx} = 0$.
$\frac{1 - \ln x}{x^2} = 0$
$1 - \ln x = 0$
$\ln x = 1$
$x = \text{e}$

3. Find the corresponding y-coordinate

When $x = \text{e}$,
$y = \frac{\ln \text{e}}{\text{e}}$
$y = \frac{1}{\text{e}}$
The coordinates are $(\text{e}, \frac{1}{\text{e}})$.

1. Find the first derivative of the curve (from part a)

From part (a), $\frac{dy}{dx} = \frac{1 - \ln x}{x^2}$.

2. Find the second derivative using the quotient rule

$\frac{d^2y}{dx^2} = \frac{x^2 \left(-\frac{1}{x}\right) - (1 - \ln x)(2x)}{(x^2)^2}$
$= \frac{-x - 2x + 2x\ln x}{x^4}$
$= \frac{x(2\ln x - 3)}{x^4}$
$= \frac{2\ln x - 3}{x^3}$

3. Evaluate the second derivative at the stationary point to determine its nature

At $x = e$,
$\frac{d^2y}{dx^2} = \frac{2\ln e - 3}{e^3} = \frac{2(1) - 3}{e^3} = -\frac{1}{e^3}$
Since $-\frac{1}{e^3} < 0$, the stationary point is a maximum.