The equation of a curve is $y = x^2(6 - x)$.

1. Expand the equation of the curve and find the first derivative

$y = x^2(6 - x) = 6x^2 - x^3$

$\frac{dy}{dx} = 12x - 3x^2$

2. Set the derivative to zero to find the $x$-coordinates of the stationary points

For stationary points, $\frac{dy}{dx} = 0$:
$12x - 3x^2 = 0$
$3x(4 - x) = 0$
$x = 0$ or $x = 4$

3. Find the corresponding $y$-coordinates

When $x = 0$, $y = 0^2(6 - 0) = 0$
When $x = 4$, $y = 4^2(6 - 4) = 16(2) = 32$

4. State the coordinates of the stationary points

The coordinates of the stationary points are $(0, 0)$ and $(4, 32)$.

1. Find the second derivative of the curve

From part (i), $\frac{dy}{dx} = 12x - 3x^2$.

$\frac{d^2y}{dx^2} = 12 - 6x$

2. Apply the second derivative test to the first stationary point

At $x = 0$:
$\frac{d^2y}{dx^2} = 12 - 6(0) = 12$
Since $12 > 0$, the point $(0, 0)$ is a minimum point.

3. Apply the second derivative test to the second stationary point

At $x = 4$:
$\frac{d^2y}{dx^2} = 12 - 6(4) = -12$
Since $-12 < 0$, the point $(4, 32)$ is a maximum point.

1. Set up the condition for the function to be increasing

For $y$ to be increasing, $\frac{dy}{dx} > 0$.
From part (i), $\frac{dy}{dx} = 12x - 3x^2$.
$12x - 3x^2 > 0$

2. Solve the quadratic inequality

$3x(4 - x) > 0$
The roots are $x = 0$ and $x = 4$. Since the coefficient of $x^2$ is negative, the expression is positive between the roots.
$0 < x < 4$