The equation of a curve is $y = xe^{-3x}$. Find the stationary point(s) and determine the nature of the stationary point(s).

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

$y = xe^{-3x}$

Let $u = x$ and $v = e^{-3x}$.
$\frac{du}{dx} = 1$
$\frac{dv}{dx} = -3e^{-3x}$

$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$
$\frac{dy}{dx} = x(-3e^{-3x}) + e^{-3x}(1)$
$\frac{dy}{dx} = e^{-3x}(1 - 3x)$

2. Equate the first derivative to zero to find the $x$-coordinate of the stationary point.

For stationary points, $\frac{dy}{dx} = 0$.
$e^{-3x}(1 - 3x) = 0$

Since $e^{-3x} > 0$ for all real $x$,
$1 - 3x = 0$
$3x = 1$
$x = \frac{1}{3}$

3. Find the corresponding $y$-coordinate.

When $x = \frac{1}{3}$,
$y = \left(\frac{1}{3}\right)e^{-3(\frac{1}{3})}$
$y = \frac{1}{3}e^{-1} = \frac{1}{3e}$

The stationary point is $\left(\frac{1}{3}, \frac{1}{3e}\right)$.

4. Find the second derivative to determine the nature of the stationary point.

$\frac{dy}{dx} = e^{-3x}(1 - 3x)$

Using the product rule again:
$\frac{d^2y}{dx^2} = e^{-3x}(-3) + (1 - 3x)(-3e^{-3x})$
$\frac{d^2y}{dx^2} = -3e^{-3x} - 3e^{-3x} + 9xe^{-3x}$
$\frac{d^2y}{dx^2} = e^{-3x}(9x - 6)$

5. Evaluate the second derivative at the stationary point to conclude its nature.

When $x = \frac{1}{3}$,
$\frac{d^2y}{dx^2} = e^{-1}\left(9\left(\frac{1}{3}\right) - 6\right)$
$\frac{d^2y}{dx^2} = e^{-1}(3 - 6) = -3e^{-1}$

Since $e^{-1} > 0$, $\frac{d^2y}{dx^2} < 0$.
Therefore, the stationary point is a maximum point.