The function $\mathrm{f}$ is defined by $\mathrm{f} : x \mapsto \frac{x-1}{x}$, $x \in \mathbb{R}$, $x \ne 0$ and $x \ne 1$.

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

Let $y = \mathrm{f}(x)$.
$y = \frac{x-1}{x}$
$xy = x - 1$
$x - xy = 1$
$x(1 - y) = 1$
$x = \frac{1}{1-y}$
Therefore, the rule is $\mathrm{f}^{-1}(x) = \frac{1}{1-x}$.

2. Determine the domain of $\mathrm{f}^{-1}$ by finding the range of $\mathrm{f}$.

The domain of $\mathrm{f}^{-1}$ is exactly the range of $\mathrm{f}$.
Rewrite $\mathrm{f}(x)$ as:
$y = 1 - \frac{1}{x}$
Since $x \in \mathbb{R}$ and $x \neq 0$, the term $\frac{1}{x} \neq 0$, which implies $y \neq 1$.
Additionally, we are given the domain restriction $x \neq 1$. Substituting $x = 1$ into $1 - \frac{1}{x}$ would give $0$, so $y \neq 0$.
Hence, the range of $\mathrm{f}$ is $\mathbb{R} \setminus \{0, 1\}$.

3. State the complete formal definition of $\mathrm{f}^{-1}$.

Combining the rule and the established domain, we state the definition in mapping notation:
$\mathrm{f}^{-1} : x \mapsto \frac{1}{1-x}, \quad x \in \mathbb{R}, x \neq 0, x \neq 1$

1. Determine the range of $\mathrm{f}$ and verify the condition for the existence of $\mathrm{f}^2$.

$\mathrm{D}_\mathrm{f} = \mathbb{R} \setminus \{0, 1\}$
Let $y = \frac{x-1}{x} \implies xy = x - 1 \implies x(1-y) = 1 \implies x = \frac{1}{1-y}$.
For $x$ to be a real number and $x \neq 0$, we require $y \neq 1$.
Additionally, since $x \neq 1$, we have $\frac{1}{1-y} \neq 1 \implies 1 - y \neq 1 \implies y \neq 0$.
Thus, the range is $\mathrm{R}_\mathrm{f} = \mathbb{R} \setminus \{0, 1\}$.
Since $\mathrm{R}_\mathrm{f} \subseteq \mathrm{D}_\mathrm{f}$ (they are exactly equal), the composite function $\mathrm{f}^2$ exists.

2. Evaluate the composite function $\mathrm{f}^2(x)$.

$\mathrm{f}^2(x) = \mathrm{f}\left(\frac{x-1}{x}\right)$
$\mathrm{f}^2(x) = \frac{\frac{x-1}{x} - 1}{\frac{x-1}{x}}$
$\mathrm{f}^2(x) = \frac{\frac{x - 1 - x}{x}}{\frac{x-1}{x}}$
$\mathrm{f}^2(x) = \frac{-1}{x-1} = \frac{1}{1-x}$

3. Compare $\mathrm{f}^2(x)$ with $\mathrm{f}^{-1}(x)$ to complete the proof.

From part (i) or the rearrangement in Step 1 where $x = \frac{1}{1-y}$, the inverse function is:
$\mathrm{f}^{-1}(x) = \frac{1}{1-x}$
Since $\mathrm{f}^2(x) = \frac{1}{1-x}$ and $\mathrm{f}^{-1}(x) = \frac{1}{1-x}$, it follows that:
$\mathrm{f}^2(x) = \mathrm{f}^{-1}(x) \text{ (shown)}$

1. Determine $\mathrm{f}^3(x)$ using the result from part (ii)

From part (ii), we know that $\mathrm{f}^2(x) = \mathrm{f}^{-1}(x)$.
Therefore, $\mathrm{f}^3(x) = \mathrm{f}(\mathrm{f}^2(x)) = \mathrm{f}(\mathrm{f}^{-1}(x))$.
Since applying a function to its inverse yields the original input, we have:
$\mathrm{f}^3(x) = x$

1. Concisely establish the expressions for $\mathrm{f}^{-1}(x)$ and $\mathrm{f}^3(x)$ from earlier parts.

Let $y = \frac{x-1}{x} \implies xy = x - 1 \implies x(1-y) = 1 \implies x = \frac{1}{1-y}$.
Thus, $\mathrm{f}^{-1}(x) = \frac{1}{1-x}$.
From part (ii), since $\mathrm{f}^2(x) = \mathrm{f}^{-1}(x)$, we have:
$\mathrm{f}^3(x) = \mathrm{f}(\mathrm{f}^2(x)) = \mathrm{f}(\mathrm{f}^{-1}(x)) = x$.

2. Simplify the equation by finding the periodic pattern of the composite function.

Since $\mathrm{f}^3(x) = x$, applying $\mathrm{f}$ three times acts as an identity function.
$2014 = 3(671) + 1$
Therefore, $\mathrm{f}^{2014}(x) = \mathrm{f}^{3(671)+1}(x) = \mathrm{f}(x)$.
The given equation becomes $\mathrm{f}(\mathrm{g}(x)) = 1 - \mathrm{e}^{-x}$.

3. Solve for $\mathrm{g}(x)$ by applying the inverse function.

$\mathrm{g}(x) = \mathrm{f}^{-1}(1 - \mathrm{e}^{-x})$
Substituting $1 - \mathrm{e}^{-x}$ into $\mathrm{f}^{-1}(x) = \frac{1}{1-x}$:
$\mathrm{g}(x) = \frac{1}{1 - (1 - \mathrm{e}^{-x})}$
$\mathrm{g}(x) = \frac{1}{\mathrm{e}^{-x}} = \mathrm{e}^x$