[Cambridge/9709/2010]
The parametric equations of a curve are
$x = 1 + \ln(t - 2), \quad y = t + \frac{9}{t}, \quad \text{for } t > 2.$

1. Differentiate $x$ with respect to $t$.

$\frac{\text{d}x}{\text{d}t} = \frac{1}{t-2}$

2. Differentiate $y$ with respect to $t$ and combine into a single fraction.

$y = t + 9t^{-1}$
$\frac{\text{d}y}{\text{d}t} = 1 - 9t^{-2} = 1 - \frac{9}{t^2} = \frac{t^2 - 9}{t^2}$

3. Apply the chain rule for parametric differentiation to find $\frac{\text{d}y}{\text{d}x}$.

$\frac{\text{d}y}{\text{d}x} = \frac{\frac{\text{d}y}{\text{d}t}}{\frac{\text{d}x}{\text{d}t}}$
$\frac{\text{d}y}{\text{d}x} = \frac{\frac{t^2 - 9}{t^2}}{\frac{1}{t-2}}$
$\frac{\text{d}y}{\text{d}x} = \frac{(t^2 - 9)(t - 2)}{t^2}$

1. Equate the gradient to zero and solve for $t$, rejecting invalid values based on the given domain.

$\frac{\text{d}y}{\text{d}x} = 0$
$\frac{(t^2 - 9)(t - 2)}{t^2} = 0$
$(t^2 - 9)(t - 2) = 0$
$(t - 3)(t + 3)(t - 2) = 0$
Since the curve is only defined for $t > 2$, we discard $t = -3$ and $t = 2$.
The only valid solution is $t = 3$.

2. Substitute $t = 3$ into the parametric equations to find the $x$ and $y$ coordinates.

Substitute $t = 3$ into the equation for $x$:
$x = 1 + \ln(3 - 2)$
$x = 1 + \ln(1)$
$x = 1 + 0 = 1$

Substitute $t = 3$ into the equation for $y$:
$y = 3 + \frac{9}{3}$
$y = 3 + 3 = 6$

The coordinates of the point are $(1, 6)$.