The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by
$\begin{aligned} \mathrm{f} &: x \mapsto \frac{1}{x} - 1, \quad x \ne 0, \\ \mathrm{g} &: x \mapsto (x+2)^2, \quad x > -2. \end{aligned}$

1. Substitute $\mathrm{g}(x)$ into $\mathrm{f}(x)$ to construct the composite function $\mathrm{fg}(x)$.

$\mathrm{fg}(x) = \mathrm{f}(\mathrm{g}(x))$
$= \mathrm{f}((x+2)^2)$
$= \frac{1}{(x+2)^2} - 1$

2. Combine the terms into a single fraction and factorise the numerator using the difference of two squares to achieve the final shown form.

$= \frac{1 - (x+2)^2}{(x+2)^2}$
Using $a^2 - b^2 = (a-b)(a+b)$:
$= \frac{[1 - (x+2)][1 + (x+2)]}{(x+2)^2}$
$= \frac{(-x-1)(x+3)}{(x+2)^2}$
$= -\frac{(x+1)(x+3)}{(x+2)^2} \quad \text{(Shown)}$

3. Determine the domain of the composite function $\mathrm{fg}$.

The domain of a composite function $\mathrm{fg}$ is defined by $\{x \in \mathrm{D}_\mathrm{g} \mid \mathrm{g}(x) \in \mathrm{D}_\mathrm{f}\}$.
Given $\mathrm{D}_\mathrm{g} = (\text{all } x > -2)$ and $\mathrm{D}_\mathrm{f} = (\text{all } x \ne 0)$.
For any $x > -2$, $\mathrm{g}(x) = (x+2)^2 > 0$, so the condition $\mathrm{g}(x) \ne 0$ is naturally satisfied.
Therefore, the domain of $\mathrm{fg}$ is the domain of $\mathrm{g}$.
$\mathrm{D}_{\mathrm{fg}} = \{x \in \mathbb{R} : x > -2\}$

1. Set up the inequality using the expression from part (i) and find the critical values.

$\mathrm{fg}(x) \ge 0$
$-\frac{(x+1)(x+3)}{(x+2)^2} \ge 0$
$\frac{(x+1)(x+3)}{(x+2)^2} \le 0$

The critical values are $x = -1$ and $x = -3$ from the numerator, and the asymptote is $x = -2$ from the denominator.

2. Solve the rational inequality without domain restrictions.

Using a sign test or number line for the expression $\frac{(x+1)(x+3)}{(x+2)^2}$:

• For $x < -3$: expression is positive
• For $-3 < x < -2$: expression is negative
• For $-2 < x < -1$: expression is negative
• For $x > -1$: expression is positive

Thus, the solution to the algebraic inequality is:
$-3 \le x \le -1, \quad x \ne -2$

3. Apply the domain restriction of the composite function to state the final answer.

From part (i), the domain of $\mathrm{fg}$ is $x > -2$.
Taking the intersection of the general solution ($-3 \le x \le -1, x \ne -2$) and the domain ($x > -2$):
$-2 < x \le -1$