The function f is defined by $f : x \mapsto \dfrac{1}{x^2 + 6x + 5}$, for $x \ge -3$, $x \ne -1$. The function g is defined by $g : x \mapsto e^x$, for $x \in \mathbb{R}$.

1. Set up the equation $y = \mathrm{f}(x)$ and rearrange to isolate the quadratic expression in $x$.

Let $y = f(x)$.
$y = \frac{1}{x^2 + 6x + 5}$
$x^2 + 6x + 5 = \frac{1}{y}$

2. Complete the square to make $x$ the subject.

$(x+3)^2 - 9 + 5 = \frac{1}{y}$
$(x+3)^2 - 4 = \frac{1}{y}$
$(x+3)^2 = \frac{1}{y} + 4$
$x+3 = \pm\sqrt{\frac{1}{y} + 4}$
$x = -3 \pm \sqrt{\frac{1}{y} + 4}$

3. Apply the domain restriction to choose the correct root and state $f^{-1}(x)$.

Since the domain of $f$ is $x \ge -3$ (with $x \ne -1$), we have $x + 3 \ge 0$.
Thus, we must take the positive root:
$x = -3 + \sqrt{\frac{1}{y} + 4}$
Replacing $y$ with $x$ to express the inverse function:
$f^{-1}(x) = -3 + \sqrt{\frac{1}{x} + 4}$

1. Express the denominator in completed square form to easily track its values.

We first complete the square for the quadratic expression in the denominator:
$x^2 + 6x + 5 = (x + 3)^2 - 3^2 + 5 = (x + 3)^2 - 4$
Thus, $\mathrm{f}(x) = \dfrac{1}{(x + 3)^2 - 4}$.

2. Determine the range of possible values for the denominator based on the given domain restrictions.

Given the domain is $x \ge -3$ and $x \ne -1$:
$x + 3 \ge 0 \quad \text{and} \quad x + 3 \ne 2$
Squaring these conditions gives:
$(x + 3)^2 \ge 0 \quad \text{and} \quad (x + 3)^2 \ne 4$
Subtracting $4$ to match the denominator:
$(x + 3)^2 - 4 \ge -4 \quad \text{and} \quad (x + 3)^2 - 4 \ne 0$
This means the denominator takes values in the intervals $[-4, 0)$ and $(0, \infty)$.

3. Deduce the overall range by taking the reciprocal of the denominator's values.

We evaluate the reciprocal for each interval of the denominator.
For $-4 \le (x + 3)^2 - 4 < 0$, taking the reciprocal of negative values reverses the inequality for the lower bound:
$\dfrac{1}{(x + 3)^2 - 4} \le -\dfrac{1}{4}$
For $(x + 3)^2 - 4 > 0$, taking the reciprocal of positive values maintains strict positivity:
$\dfrac{1}{(x + 3)^2 - 4} > 0$
Therefore, combining these cases gives the overall range of $\mathrm{f}$.

1. Determine the range of the inner function $f$ (as carried over from part b)

$f(x) = \dfrac{1}{x^2 + 6x + 5} = \dfrac{1}{(x+3)^2 - 4}$
Given the domain $x \ge -3$ and $x \ne -1$, we have $(x+3)^2 \ge 0$ and $(x+3)^2 \ne 4$.
The denominator $(x+3)^2 - 4$ therefore takes values in $[-4, 0) \cup (0, \infty)$.
Taking the reciprocal, the range of $f$ is $\text{R}_f = (-\infty, -\frac{1}{4}] \cup (0, \infty)$.

2. Evaluate the range of the composite function $gf$ by applying $g$ to the range of $f$

The range of $gf$ is the set of values obtained by substituting $\text{R}_f$ into the outer function $g(x) = e^x$.
Since $g(x) = e^x$ is a strictly increasing, continuous function for all real $x$:
For $y \in (-\infty, -\frac{1}{4}]$, $0 < e^y \le e^{-1/4}$.
For $y \in (0, \infty)$, $e^y > e^0 = 1$.
Combining these intervals, the exact range of $gf$ is $(0, e^{-1/4}] \cup (1, \infty)$.