一条曲线的方程为 $y = \dfrac{\ln 2x}{x}$。

1. Find the first derivative $\frac{dy}{dx}$ using the quotient rule.

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

2. Find the second derivative $\frac{d^{2}y}{dx^{2}}$ using the quotient rule again.

Let $u = 1 - \ln 2x$ and $v = x^{2}$.
$\frac{du}{dx} = -\frac{1}{x}$
$\frac{dv}{dx} = 2x$
$\frac{d^{2}y}{dx^{2}} = \frac{x^{2}(-\frac{1}{x}) - (1 - \ln 2x)(2x)}{(x^{2})^{2}}$
$= \frac{-x - 2x + 2x\ln 2x}{x^{4}}$
$= \frac{-3x + 2x\ln 2x}{x^{4}}$
$= \frac{2\ln 2x - 3}{x^{3}}$

3. Substitute $\frac{dy}{dx}$ and $\frac{d^{2}y}{dx^{2}}$ into the given expression to show it equals 0.

LHS $= x\left(\frac{2\ln 2x - 3}{x^{3}}\right) + 2\left(\frac{1 - \ln 2x}{x^{2}}\right) + \frac{1}{x^{2}}$
$= \frac{2\ln 2x - 3}{x^{2}} + \frac{2 - 2\ln 2x}{x^{2}} + \frac{1}{x^{2}}$
$= \frac{2\ln 2x - 3 + 2 - 2\ln 2x + 1}{x^{2}}$
$= \frac{0}{x^{2}} = 0 = \text{RHS (shown)}$

1. Find the first derivative $\frac{dy}{dx}$ and set it to zero to find the stationary point.

$y = \frac{\ln 2x}{x}$

Using the quotient rule:
$\frac{dy}{dx} = \frac{x \left(\frac{1}{2x} \cdot 2\right) - (\ln 2x)(1)}{x^2} = \frac{1 - \ln 2x}{x^2}$

For a stationary point, $\frac{dy}{dx} = 0$:
$\frac{1 - \ln 2x}{x^2} = 0$

2. Solve for the exact $x$-coordinate of the stationary point.

$1 - \ln 2x = 0$
$\ln 2x = 1$
$2x = e$
$x = \frac{e}{2}$

3. Find the exact $y$-coordinate of the stationary point.

Substitute $x = \frac{e}{2}$ into the original equation:
$y = \frac{\ln(2 \cdot \frac{e}{2})}{\frac{e}{2}} = \frac{\ln e}{\frac{e}{2}} = \frac{1}{\frac{e}{2}} = \frac{2}{e}$

The exact coordinates are $\left(\frac{e}{2}, \frac{2}{e}\right)$.

4. Use the relation from part (i) to evaluate the second derivative at the stationary point.

From part (i), we have $x\frac{d^{2}y}{dx^{2}} + 2\frac{dy}{dx} + \frac{1}{x^{2}} = 0$.
At the stationary point, $\frac{dy}{dx} = 0$, so:
$x\frac{d^{2}y}{dx^{2}} + 0 + \frac{1}{x^{2}} = 0$
$\frac{d^{2}y}{dx^{2}} = -\frac{1}{x^{3}}$

Substitute $x = \frac{e}{2}$:
$\frac{d^{2}y}{dx^{2}} = -\frac{1}{(\frac{e}{2})^{3}} = -\frac{8}{e^{3}}$

5. Determine the nature of the stationary point.

Since $e > 0$, $-\frac{8}{e^{3}} < 0$.
Because $\frac{d^{2}y}{dx^{2}} < 0$, the stationary point is a maximum.