函数 $f$ 定义为 $f : x \mapsto x^2 - 4x$，$x \in \mathbb{R}$，$|x| < 1$。

1. Show that the function is one-to-one to justify the existence of its inverse.

$f(x) = x^2 - 4x = (x-2)^2 - 4$
The graph of $y = f(x)$ is a parabola opening upwards with a line of symmetry at $x = 2$.
Since the given domain is $|x| < 1$ (i.e. $-1 < x < 1$), all possible values of $x$ satisfy $x < 2$.
Therefore, $f(x)$ is strictly decreasing on its entire domain. Since it is strictly monotonic, $f$ is a one-to-one function, which means $f^{-1}$ exists.

2. Find the rule for the inverse function by making $x$ the subject.

Let $y = f(x) = (x-2)^2 - 4$.
$(x-2)^2 = y + 4$
$x - 2 = \pm\sqrt{y+4}$
Since $x \in (-1, 1)$, we know $x - 2 < -1 < 0$. Therefore, we must reject the positive square root:
$x - 2 = -\sqrt{y+4}$
$x = 2 - \sqrt{y+4}$
Thus, $f^{-1}(x) = 2 - \sqrt{x+4}$.

3. Determine the domain of the inverse function.

The domain of $f^{-1}$ is exactly the range of $f$.
Since $f(x)$ is strictly decreasing on $-1 < x < 1$, we can find its range by evaluating the limits at the endpoints:
When $x \to -1$, $f(x) \to (-1)^2 - 4(-1) = 5$.
When $x \to 1$, $f(x) \to (1)^2 - 4(1) = -3$.
Thus, the range of $f$ is $-3 < f(x) < 5$, meaning the domain of $f^{-1}$ is $-3 < x < 5$.

4. Express the inverse function in the required similar mapping notation.

Expressing $f^{-1}$ in a similar form to the definition of $f$:
$f^{-1} : x \mapsto 2 - \sqrt{x+4}, \; x \in \mathbb{R}, \; -3 < x < 5$

1. Find the range of the inner function $f$.

$f(x) = x^2 - 4x = (x-2)^2 - 4$
For the domain $-1 < x < 1$, the function is strictly decreasing since its turning point is at $x=2$ (which is outside the domain).
As $x \to -1^+$, $f(x) \to (-1)^2 - 4(-1) = 5$.
As $x \to 1^-$, $f(x) \to (1)^2 - 4(1) = -3$.
Therefore, the range of $f$ is $R_f = (-3, 5)$.

2. Compare $R_f$ with the domain of $g$ to conclude the existence of $gf$.

The domain of $g$ is given as $x \ne -3$, so $D_g = \mathbb{R} \setminus \{-3\}$.
Since $-3$ is not included in the open interval $(-3, 5)$, all values in $R_f$ are valid inputs for $g$.
Thus, $R_f \subseteq D_g$.
Hence, the composite function $gf$ exists.

1. Substitute the expression for $f(x)$ into $g(x)$

$gf(x) = g(x^2 - 4x)$
$gf(x) = 1 + \dfrac{1}{(x^2 - 4x) + 3}$
$= 1 + \dfrac{1}{x^2 - 4x + 3}$

1. Find the first derivative of $h(x)$ using the expression derived in part (iii).

From part (iii), $h(x) = gf(x) = 1 + (x^2 - 4x + 3)^{-1}$.
Using the chain rule to differentiate $h$ with respect to $x$:
$h'(x) = -(x^2 - 4x + 3)^{-2}(2x - 4)$
$h'(x) = \frac{4 - 2x}{(x^2 - 4x + 3)^2}$

2. Determine the sign of the derivative for the given domain.

The domain is given as $|x| < 1$, which means $-1 < x < 1$.
For $-1 < x < 1$, $x < 2 \implies 4 - 2x > 0$.
The denominator $(x^2 - 4x + 3)^2 > 0$ for all $x$ in the domain.
Therefore, $h'(x) > 0$ for all $x \in (-1, 1)$.

3. Conclude whether the function is increasing or decreasing.

Since $h'(x) > 0$ across the entire domain $(-1, 1)$, $h$ is strictly increasing.

4. Evaluate the limits at the endpoints of the domain to find the range.

Because $h$ is strictly increasing, its range is bounded by its limits as $x$ approaches the domain endpoints $-1$ and $1$.
As $x \to -1^+$, $h(x) \to h(-1) = 1 + \frac{1}{(-1)^2 - 4(-1) + 3} = 1 + \frac{1}{8} = \frac{9}{8}$.
As $x \to 1^-$, the expression $x^2 - 4x + 3 = (x-1)(x-3) \to 0^+$.
Hence, $\frac{1}{x^2 - 4x + 3} \to \infty$, which means $h(x) \to \infty$.
Thus, the range of $h$ is $\left( \frac{9}{8}, \infty \right)$.