已知 $y = \mathrm{e}^{-x}\ln(1-2x)$。

1. Find the first derivative of $y$ with respect to $x$ using the product rule.

Given $y = \mathrm{e}^{-x}\ln(1-2x)$, differentiate with respect to $x$:
$\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{-x} \left( \frac{-2}{1-2x} \right) + \ln(1-2x) \left( -\mathrm{e}^{-x} \right)$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-2\mathrm{e}^{-x}}{1-2x} - \mathrm{e}^{-x}\ln(1-2x)$

2. Multiply by $(1-2x)$ and substitute $y$ back into the equation to show the first target.

Multiply the entire equation by $(1-2x)$:
$(1-2x)\frac{\mathrm{d}y}{\mathrm{d}x} = -2\mathrm{e}^{-x} - (1-2x)\mathrm{e}^{-x}\ln(1-2x)$
Since $y = \mathrm{e}^{-x}\ln(1-2x)$, substitute $y$ back into the right side:
$(1-2x)\frac{\mathrm{d}y}{\mathrm{d}x} = -2\mathrm{e}^{-x} - y(1-2x) \quad \text{(Shown)}$

3. Differentiate the new implicit equation with respect to $x$.

Differentiate $(1-2x)\frac{\mathrm{d}y}{\mathrm{d}x} = -2\mathrm{e}^{-x} - y(1-2x)$ implicitly with respect to $x$. Applying the product rule to both $(1-2x)\frac{\mathrm{d}y}{\mathrm{d}x}$ and $y(1-2x)$:
$(1-2x)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + (-2)\frac{\mathrm{d}y}{\mathrm{d}x} = 2\mathrm{e}^{-x} - \left( y(-2) + (1-2x)\frac{\mathrm{d}y}{\mathrm{d}x} \right)$
$(1-2x)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2\frac{\mathrm{d}y}{\mathrm{d}x} = 2\mathrm{e}^{-x} + 2y - (1-2x)\frac{\mathrm{d}y}{\mathrm{d}x}$

4. Substitute the first target result back in and rearrange to match the final target.

Substitute $(1-2x)\frac{\mathrm{d}y}{\mathrm{d}x} = -2\mathrm{e}^{-x} - y(1-2x)$ into the right-hand side to simplify the expression:
$(1-2x)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2\frac{\mathrm{d}y}{\mathrm{d}x} = 2\mathrm{e}^{-x} + 2y - \left( -2\mathrm{e}^{-x} - y(1-2x) \right)$
$(1-2x)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2\frac{\mathrm{d}y}{\mathrm{d}x} = 2\mathrm{e}^{-x} + 2y + 2\mathrm{e}^{-x} + y(1-2x)$
Group and simplify the terms on the right:
$(1-2x)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2\frac{\mathrm{d}y}{\mathrm{d}x} = 4\mathrm{e}^{-x} + 2y + y - 2xy$
$(1-2x)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2\frac{\mathrm{d}y}{\mathrm{d}x} = 4\mathrm{e}^{-x} + (3-2x)y$
Rearrange to isolate $(1-2x)\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$:
$(1-2x)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2\frac{\mathrm{d}y}{\mathrm{d}x} + 4\mathrm{e}^{-x} + (3-2x)y \quad \text{(Shown)}$

1. Evaluate the function and its first two derivatives at $x=0$ using the given equations.

$y(0) = \mathrm{e}^{0}\ln(1) = 0$

Substitute $x=0, y=0$ into the first equation $(1-2x)\frac{\mathrm{d}y}{\mathrm{d}x} = -2\mathrm{e}^{-x} - y(1-2x)$:
$1 \cdot y'(0) = -2(1) - 0 \implies y'(0) = -2$

Substitute $x=0, y=0, y'=-2$ into the second equation $(1-2x)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2\frac{\mathrm{d}y}{\mathrm{d}x} + 4\mathrm{e}^{-x} + (3-2x)y$:
$1 \cdot y''(0) = 2(-2) + 4(1) + 3(0) = -4 + 4 \implies y''(0) = 0$

2. Differentiate the second derivative equation implicitly with respect to $x$ to find an expression for the third derivative.

Differentiating both sides of $(1-2x)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2\frac{\mathrm{d}y}{\mathrm{d}x} + 4\mathrm{e}^{-x} + (3-2x)y$ with respect to $x$ using the product rule:
$(1-2x)\frac{\mathrm{d}^3y}{\mathrm{d}x^3} - 2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 4\mathrm{e}^{-x} + (3-2x)\frac{\mathrm{d}y}{\mathrm{d}x} - 2y$

3. Evaluate the third derivative at $x=0$.

Substitute $x=0, y=0, y'=-2$, and $y''=0$ into the new equation:
$(1)y'''(0) - 2(0) = 2(0) - 4(1) + 3(-2) - 2(0)$
$y'''(0) = -4 - 6 = -10$

4. Construct the Maclaurin series up to the $x^3$ term.

Using the Maclaurin series formula $y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots$
$y = 0 + (-2)x + \frac{0}{2!}x^2 + \frac{-10}{3!}x^3 + \dots$
$y = -2x - \frac{10}{6}x^3$
$y \approx -2x - \frac{5}{3}x^3$

1. Concisely obtain the result from part (ii) by evaluating the derivatives at $x=0$, as it is required for verification.

When $x = 0$, $y = \mathrm{e}^{0}\ln(1) = 0$.
From part (i), $(1-2x)\frac{\mathrm{d}y}{\mathrm{d}x} = -2\mathrm{e}^{-x} - y(1-2x)$. Substituting $x=0$ gives $1 \cdot y'(0) = -2 - 0 \implies y'(0) = -2$.
Also from part (i), $(1-2x)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2\frac{\mathrm{d}y}{\mathrm{d}x} + 4\mathrm{e}^{-x} + (3-2x)y$.
Substituting $x=0$ gives $1 \cdot y''(0) = 2(-2) + 4 + 0 \implies y''(0) = 0$.
Differentiating the second equation with respect to $x$:
$(1-2x)\frac{\mathrm{d}^3y}{\mathrm{d}x^3} - 2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 4\mathrm{e}^{-x} + (3-2x)\frac{\mathrm{d}y}{\mathrm{d}x} - 2y$
Substituting $x=0$ gives $1 \cdot y'''(0) - 0 = 0 - 4 + 3(-2) - 0 \implies y'''(0) = -10$.
Hence, the series from part (ii) is:
$y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 = -2x - \frac{10}{6}x^3 = -2x - \frac{5}{3}x^3$

2. Write out the standard Maclaurin series expansions for $\mathrm{e}^{-x}$ and $\ln(1-2x)$ up to the term in $x^3$.

Using the standard series $\mathrm{e}^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$ and $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$:

$\mathrm{e}^{-x} = 1 + (-x) + \frac{(-x)^2}{2!} + \frac{(-x)^3}{3!} + \dots$
$\mathrm{e}^{-x} = 1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3 + \dots$

$\ln(1-2x) = (-2x) - \frac{(-2x)^2}{2} + \frac{(-2x)^3}{3} - \dots$
$\ln(1-2x) = -2x - \frac{4x^2}{2} - \frac{8x^3}{3} - \dots = -2x - 2x^2 - \frac{8}{3}x^3 - \dots$

3. Multiply the two series expansions up to the $x^3$ term to verify the result.

$y = \left( 1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3 + \dots \right)\left( -2x - 2x^2 - \frac{8}{3}x^3 - \dots \right)$
Expanding and ignoring terms with powers of $x$ greater than 3:
$y \approx 1\left( -2x - 2x^2 - \frac{8}{3}x^3 \right) - x\left( -2x - 2x^2 \right) + \frac{1}{2}x^2\left( -2x \right)$
$y \approx -2x - 2x^2 - \frac{8}{3}x^3 + 2x^2 + 2x^3 - x^3$
$y \approx -2x + \left( -2 + 2 \right)x^2 + \left( -\frac{8}{3} + 2 - 1 \right)x^3$
$y \approx -2x - \frac{5}{3}x^3$
This is identically the result obtained in part (ii), verifying the standard series expansions yield the same result.