The gradient function of a curve is given by $\frac{\text{d}y}{\text{d}x} = \frac{16}{x^3} - 4$. Given that the line $2y - x - 6 = 0$ is a normal to the curve, find the equation of the curve.

1. Find the gradient of the normal and the gradient of the tangent

The equation of the normal is $2y - x - 6 = 0$, which can be rewritten as $y = \frac{1}{2}x + 3$.
The gradient of the normal is $m_n = \frac{1}{2}$.
Since the tangent is perpendicular to the normal, the gradient of the tangent is $m_t = -\frac{1}{m_n} = -2$.

2. Find the $x$-coordinate of the point on the curve

Equate the gradient function to the gradient of the tangent:
$\frac{16}{x^3} - 4 = -2$
$\frac{16}{x^3} = 2$
$x^3 = 8$
$x = 2$

3. Find the $y$-coordinate of the point on the curve

Since the normal passes through the point of tangency, substitute $x = 2$ into the equation of the normal:
$y = \frac{1}{2}(2) + 3 = 4$
The point on the curve is $(2, 4)$.

4. Integrate the gradient function to find the general equation of the curve

$y = \int \left( \frac{16}{x^3} - 4 \right) \text{d}x$
$y = \int \left( 16x^{-3} - 4 \right) \text{d}x$
$y = \frac{16x^{-2}}{-2} - 4x + c$
$y = -\frac{8}{x^2} - 4x + c$

5. Find the constant of integration and state the final equation

Substitute the point $(2, 4)$ into the general equation:
$4 = -\frac{8}{2^2} - 4(2) + c$
$4 = -2 - 8 + c$
$4 = -10 + c \implies c = 14$

The equation of the curve is $y = -\frac{8}{x^2} - 4x + 14$.