A curve is such that $\frac{dy}{dx} = \frac{8}{x^{2}} - 2.$

1. Integrate the given derivative to find the general equation of the curve

$\frac{dy}{dx} = 8x^{-2} - 2$
$y = \int (8x^{-2} - 2) \, dx$
$y = \frac{8x^{-1}}{-1} - 2x + c$
$y = -\frac{8}{x} - 2x + c$

2. Substitute the given point to find the constant of integration

Substitute $x = 1, y = 5$:
$5 = -\frac{8}{1} - 2(1) + c$
$5 = -10 + c$
$c = 15$

3. State the final equation of the curve

$y = -\frac{8}{x} - 2x + 15$

1. Set the derivative to zero to find stationary points

For stationary points, $\frac{dy}{dx} = 0$.
$\frac{8}{x^{2}} - 2 = 0$

2. Solve the equation for $x$

$\frac{8}{x^{2}} = 2$
$2x^{2} = 8$
$x^{2} = 4$
$x = 2 \text{ or } x = -2$

1. Differentiate $\frac{dy}{dx}$ to find $\frac{d^{2}y}{dx^{2}}$.

$\frac{dy}{dx} = 8x^{-2} - 2$
$\frac{d^{2}y}{dx^{2}} = -16x^{-3} = -\frac{16}{x^{3}}$

2. Find the stationary points from part (ii) to test.

From (ii), $\frac{dy}{dx} = 0 \implies \frac{8}{x^2} - 2 = 0 \implies x^2 = 4 \implies x = 2$ or $x = -2$.

3. Evaluate the second derivative at $x = 2$ to determine its nature.

When $x = 2$, $\frac{d^{2}y}{dx^{2}} = -\frac{16}{2^{3}} = -2 < 0$.
Therefore, the stationary point at $x = 2$ is a maximum.

4. Evaluate the second derivative at $x = -2$ to determine its nature.

When $x = -2$, $\frac{d^{2}y}{dx^{2}} = -\frac{16}{(-2)^{3}} = 2 > 0$.
Therefore, the stationary point at $x = -2$ is a minimum.