The equation of a curve is $y = \frac{x}{e^{2x}}$.

1. Apply the quotient rule to differentiate the function

Let $u = x$ and $v = e^{2x}$.
Then $\frac{du}{dx} = 1$ and $\frac{dv}{dx} = 2e^{2x}$.

$\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$
$\frac{dy}{dx} = \frac{e^{2x}(1) - x(2e^{2x})}{(e^{2x})^2}$

2. Simplify the derivative expression

$\frac{dy}{dx} = \frac{e^{2x}(1 - 2x)}{e^{4x}}$
$\frac{dy}{dx} = \frac{1 - 2x}{e^{2x}}$

1. Evaluate the derivative $\frac{dy}{dx}$ at $x = 1$

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

2. Use the chain rule to find the rate of change of $y$

$\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$
$\frac{dy}{dt} = -\frac{1}{e^2} \times 0.18$
$\frac{dy}{dt} = -\frac{0.18}{e^2} \approx -0.0244$ (to 3 s.f.)

1. Set the derivative from part (a) to zero to find stationary points

For stationary points, $\frac{dy}{dx} = 0$.
From part (a), $\frac{dy}{dx} = e^{-2x}(1 - 2x)$.
$e^{-2x}(1 - 2x) = 0$

2. Solve the equation and explain the number of solutions

Since $e^{-2x} > 0$ for all real values of $x$, $e^{-2x} \neq 0$.
Therefore, $1 - 2x = 0 \implies x = 0.5$.
Since there is only one solution for $x$, the curve has exactly one stationary point.