It is given that $y = \ln(2 - \text{e}^{-2x})$.

1. Rewrite the equation in exponential form to prepare for implicit differentiation without relying on fractions.

$y = \ln(2 - \text{e}^{-2x})$
$\text{e}^y = 2 - \text{e}^{-2x}$

2. Differentiate both sides implicitly with respect to $x$.

$\text{e}^y \frac{\text{d}y}{\text{d}x} = -(-2)\text{e}^{-2x}$
$\text{e}^y \frac{\text{d}y}{\text{d}x} = 2\text{e}^{-2x}$

3. Substitute $\text{e}^{-2x}$ in terms of $y$ and rearrange to obtain the required expression.

From the first step, we know that $\text{e}^{-2x} = 2 - \text{e}^y$. Substituting this in:
$\text{e}^y \frac{\text{d}y}{\text{d}x} = 2(2 - \text{e}^y)$
$\text{e}^y \frac{\text{d}y}{\text{d}x} = 4 - 2\text{e}^y$
Divide both sides by $\text{e}^y$:
$\frac{\text{d}y}{\text{d}x} = \frac{4}{\text{e}^y} - \frac{2\text{e}^y}{\text{e}^y}$
$\frac{\text{d}y}{\text{d}x} = 4\text{e}^{-y} - 2 \quad \text{(Shown)}$

1. Evaluate $y$ and $\frac{\mathrm{d}y}{\mathrm{d}x}$ at $x = 0$.

Substitute $x = 0$ into the original equation:
$y = \ln(2 - \mathrm{e}^{0}) = \ln(1) = 0$

Substitute $y = 0$ into the result from part (a):
$\frac{\mathrm{d}y}{\mathrm{d}x} = 4\mathrm{e}^{0} - 2 = 2$

2. Differentiate $\frac{\mathrm{d}y}{\mathrm{d}x}$ implicitly to find the second derivative.

$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{\mathrm{d}}{\mathrm{d}x} \left( 4\mathrm{e}^{-y} - 2 \right)$
$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -4\mathrm{e}^{-y} \frac{\mathrm{d}y}{\mathrm{d}x}$

3. Evaluate the second derivative at $x = 0$ and construct the Maclaurin series.

At $x = 0, y = 0, \frac{\mathrm{d}y}{\mathrm{d}x} = 2$:
$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -4(1)(2) = -8$

The Maclaurin series is given by:
$y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \ldots$
$y = 0 + 2x + \frac{-8}{2}x^2$
$y = 2x - 4x^2$

1. Expand $\text{e}^{-2x}$ up to the term in $x^2$ using the standard series.

$\text{e}^{-2x} = 1 + (-2x) + \frac{(-2x)^2}{2!} + \dots \approx 1 - 2x + 2x^2$

2. Substitute the expansion of $\text{e}^{-2x}$ into $2 - \text{e}^{-2x}$.

$2 - \text{e}^{-2x} \approx 2 - (1 - 2x + 2x^2) = 1 + 2x - 2x^2$

3. Expand $\ln(1 + u)$ where $u = 2x - 2x^2$, up to the term in $x^2$.

$\ln(2 - \text{e}^{-2x}) \approx \ln(1 + (2x - 2x^2))$

Using the standard series $\ln(1+u) = u - \frac{u^2}{2} + \dots$:
$= (2x - 2x^2) - \frac{1}{2}(2x - 2x^2)^2$
$= 2x - 2x^2 - \frac{1}{2}(4x^2 - 8x^3 + \dots)$
Discarding terms with powers higher than $x^2$:
$= 2x - 2x^2 - 2x^2$
$= 2x - 4x^2$

4. Compare the obtained expansion with the series found in part (b) to verify correctness.

In part (b), evaluating $y$, $y'$, and $y''$ at $x=0$ yields $0$, $2$, and $-8$ respectively, giving the Maclaurin series $y \approx 0 + 2x + \frac{-8}{2}x^2 = 2x - 4x^2$.
Since the expansion using standard series is also $2x - 4x^2$, it confirms the correctness of the series found in part (b).