已知 $y = \ln\left(2 - e^{-2x}\right).$

1. Rewrite the equation in exponential form and differentiate implicitly with respect to $x$.

$y = \ln\left(2 - e^{-2x}\right) \implies e^y = 2 - e^{-2x}$

Differentiating implicitly with respect to $x$:
$e^y \frac{dy}{dx} = 2e^{-2x}$

2. Express the derivative in terms of $y$ and simplify to obtain the required result.

From $e^y = 2 - e^{-2x}$, we can express $e^{-2x}$ as $2 - e^y$.
Substituting this into the differentiated equation:
$e^y \frac{dy}{dx} = 2(2 - e^y)$
$e^y \frac{dy}{dx} = 4 - 2e^y$
Dividing both sides by $e^y$:
$\frac{dy}{dx} = \frac{4 - 2e^y}{e^y}$
$\frac{dy}{dx} = 4e^{-y} - 2 \text{ (Shown)}$

1. Evaluate $y$ and the first derivative $\frac{dy}{dx}$ at $x=0$.

At $x = 0$,
$y(0) = \ln(2 - e^0) = \ln(1) = 0$

Using the result from part (a),
$\frac{dy}{dx}\bigg|_{x=0} = 4e^0 - 2 = 2$

2. Differentiate implicitly to find the second derivative $\frac{d^2y}{dx^2}$ and evaluate it at $x=0$.

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

Substituting $y=0$ and $\frac{dy}{dx}=2$,
$\frac{d^2y}{dx^2}\bigg|_{x=0} = -4(e^0)(2) = -8$

3. Apply the Maclaurin series expansion formula up to the $x^2$ term.

The Maclaurin series expansion is:
$y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \ldots$

Substituting the evaluated values:
$y = 0 + 2x + \frac{-8}{2}x^2$
$y = 2x - 4x^2$

1. Establish the result from part (b) that needs to be checked.

From part (a), $\frac{dy}{dx} = 4e^{-y} - 2$.
At $x = 0$, $y = \ln(2 - e^0) = 0$.
$\frac{dy}{dx} \Big|_{x=0} = 4(1) - 2 = 2$.
Differentiating implicitly, $\frac{d^2y}{dx^2} = -4e^{-y}\frac{dy}{dx}$.
$\frac{d^2y}{dx^2} \Big|_{x=0} = -4(1)(2) = -8$.
The Maclaurin series from part (b) is:
$y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 = 2x - 4x^2$.

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

Using the standard series $e^x = 1 + x + \frac{x^2}{2!} + \dots$,
$e^{-2x} = 1 + (-2x) + \frac{(-2x)^2}{2!} + \dots$
$e^{-2x} \approx 1 - 2x + 2x^2$

3. Substitute the expansion into the logarithmic function.

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

4. Expand $\ln(1+u)$ using the standard series, where $u = 2x - 2x^2$.

Using the standard series $\ln(1+u) = u - \frac{u^2}{2} + \dots$, substitute $u = 2x - 2x^2$:
$y \approx (2x - 2x^2) - \frac{1}{2}(2x - 2x^2)^2$

5. Expand and simplify up to the $x^2$ term, and conclude the check.

Expanding the squared term up to $x^2$ gives $(2x)^2 = 4x^2$.
$y \approx 2x - 2x^2 - \frac{1}{2}(4x^2)$
$y \approx 2x - 2x^2 - 2x^2$
$y \approx 2x - 4x^2$

This exactly matches the Maclaurin series derived in part (b), confirming its correctness.