The function f is defined as follows. f : $x -> \displaystyle -\frac{1}{x^{2}+1}$ for $x \in \mathbb{R}$.

1. Identify the $y$-intercept / turning point and horizontal asymptote.

When $x = 0$, $y = -\frac{1}{0^2+1} = -1$. Since $x^2 \ge 0$, the minimum value of $x^2+1$ is 1, so the minimum value of $f(x)$ is $-1$.
As $x \to \pm \infty$, $x^2+1 \to \infty$, which means $f(x) \to 0$.
Thus, the horizontal asymptote is $y = 0$.

2. Sketch the curve incorporating all key features.

Draw an inverted bell-shaped curve passing through a minimum point at $(0, -1)$. Ensure the curve is symmetrical about the $y$-axis and approaches $y = 0$ from below as $x \to \pm\infty$.

1. State the mathematical requirement for an inverse function to exist.

For the inverse function $f^{-1}(x)$ to exist, the function $f(x)$ must be a one-to-one function.

2. Determine the greatest value of k based on the function's turning point.

From the properties of $f(x) = -\frac{1}{x^{2}+1}$ (or its graph), there is a turning point (minimum) at $x = 0$ and the curve is symmetrical about the $y$-axis.
To ensure any horizontal line $y=c$ intersects the graph at most once (Horizontal Line Test), the restricted domain $x \le k$ must not contain points on both sides of the turning point.
Therefore, the greatest value of $k$ is $0$.

1. Deduce the restricted domain from part (ii) to use for finding the inverse function.

From part (ii), the graph of $\mathrm{f}(x) = -\frac{1}{x^2+1}$ is a curve with a maximum turning point at $x = 0$.
For $\mathrm{f}(x)$ to be a one-to-one function on $x \le k$, the domain must not cross the turning point. The greatest possible value is $k = 0$.
Thus, the restricted domain we must use is $x \le 0$.

2. Find an expression for $\mathrm{f}^{-1}(x)$ by changing the subject of the formula.

Let $y = \mathrm{f}(x)$.
$y = -\frac{1}{x^2+1}$
$x^2+1 = -\frac{1}{y}$
$x^2 = -\frac{1}{y} - 1 = -\frac{1+y}{y}$

3. Apply the restricted domain to determine the correct sign for the inverse function.

Since the restricted domain is $x \le 0$, $x$ must be non-positive. We take the negative square root:
$x = -\sqrt{-\frac{1+y}{y}}$
Therefore, an expression for $\mathrm{f}^{-1}(x)$ is:
$\mathrm{f}^{-1}(x) = -\sqrt{-\frac{1+x}{x}}$

4. Set up an equation to find the real value of $a$ such that $\mathrm{f}^{-1}(a) = \mathrm{f}(a)$.

The condition $\mathrm{f}^{-1}(a) = \mathrm{f}(a)$ means the graphs of $\mathrm{f}$ and $\mathrm{f}^{-1}$ intersect. The intersection point of a function and its inverse lies on the line $y = x$, so we can set $\mathrm{f}(a) = a$:
$-\frac{1}{a^2+1} = a$
$a(a^2+1) = -1$
$a^3 + a + 1 = 0$
(Note: Solving $\mathrm{f}^{-1}(a) = \mathrm{f}(a)$ directly with a graphing calculator is also mathematically acceptable.)

5. Solve the equation to find $a$.

Using a graphing calculator to find the real root of $a^3 + a + 1 = 0$:
$a \approx -0.68233$
$a = -0.682 \text{ (to 3 s.f.)}$

1. Determine the domain of $f^{-1}$ using the expression found in part (iii).

From part (iii), the inverse function is $f^{-1}(x) = -\sqrt{-\frac{x+1}{x}}$.
For $f^{-1}(x)$ to be mathematically defined, the expression inside the square root must be non-negative:
$-\frac{x+1}{x} \ge 0 \implies \frac{x+1}{x} \le 0$
Solving this rational inequality (with critical values $x=-1$ and $x=0$, where $x=0$ is undefined) yields the domain of $f^{-1}$:
$D_{f^{-1}} = [-1, 0)$

2. State the condition for the composite function $f^{-1}g$ to exist and set up the corresponding inequality.

For the composite function $f^{-1}g$ to exist, the range of the inner function $g$ must be a subset of the domain of the outer function $f^{-1}$:
$R_g \subseteq D_{f^{-1}}$
This means the outputs of $g(x)$ must lie within $[-1, 0)$:
$-1 \le g(x) < 0$

3. Solve the inequality to find the restriction on the domain of $g$.

Substitute $g(x) = x + 3$ into the inequality:
$-1 \le x + 3 < 0$
Subtract $3$ from all parts:
$-4 \le x < -3$
So, the maximum valid restriction on the domain of $g$ is $-4 \le x < -3$.