已知 $y = \ln(\cos ax - \sin ax)$，其中 $a$ 是一个非零常数。

1. Rewrite the function to eliminate the natural logarithm and differentiate implicitly with respect to $x$.

$y = \ln(\cos ax - \sin ax)$
$e^y = \cos ax - \sin ax$

Differentiating both sides with respect to $x$:
$e^y \frac{\mathrm{d}y}{\mathrm{d}x} = -a\sin ax - a\cos ax$

2. Differentiate implicitly a second time using the product rule.

Differentiating both sides with respect to $x$ again:
$\frac{\mathrm{d}}{\mathrm{d}x} \left( e^y \frac{\mathrm{d}y}{\mathrm{d}x} \right) = \frac{\mathrm{d}}{\mathrm{d}x} (-a\sin ax - a\cos ax)$

Applying the product rule on the left hand side:
$e^y \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right) \left(e^y \frac{\mathrm{d}y}{\mathrm{d}x}\right) = -a^2\cos ax + a^2\sin ax$
$e^y \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + e^y \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = -a^2(\cos ax - \sin ax)$

3. Substitute $e^y$ back into the equation to show the required result.

Since $e^y = \cos ax - \sin ax$, substitute this back into the right-hand side:
$e^y \left[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 \right] = -a^2 e^y$

Since $e^y > 0$ for all real $y$, we can divide both sides by $e^y$:
$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = -a^2$
$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + a^2 = 0 \quad \text{(shown)}$

1. Evaluate $y(0)$ and $y'(0)$ using the original equation.

Given $y = \ln(\cos ax - \sin ax)$,
when $x = 0$, $y(0) = \ln(1 - 0) = 0$.

Differentiating with respect to $x$:
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-a\sin ax - a\cos ax}{\cos ax - \sin ax}$
When $x = 0$,
$y'(0) = \frac{-a(0) - a(1)}{1 - 0} = -a$

2. Differentiate the equation from part (i) implicitly with respect to $x$.

Differentiating the given result $\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + a^2 = 0$ implicitly with respect to $x$:
$\frac{\mathrm{d}^3 y}{\mathrm{d}x^3} + 2\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)\left(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}\right) = 0$

3. Evaluate $y''(0)$ and $y'''(0)$.

From the result in (i), substituting $x = 0$:
$y''(0) + (-a)^2 + a^2 = 0 \implies y''(0) = -2a^2$

Substituting $x = 0$ into the differentiated equation from Step 2:
$y'''(0) + 2(-a)(-2a^2) = 0$
$y'''(0) + 4a^3 = 0 \implies y'''(0) = -4a^3$

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

The Maclaurin series is given by:
$y = y(0) + x y'(0) + \frac{x^2}{2!} y''(0) + \frac{x^3}{3!} y'''(0) + \ldots$
$y = 0 + x(-a) + \frac{x^2}{2}(-2a^2) + \frac{x^3}{6}(-4a^3) + \ldots$
$y = -ax - a^2x^2 - \frac{2}{3}a^3x^3 + \ldots$

1. Concisely obtain the Maclaurin series for $y$ up to the term in $x^3$ from part (ii).

$y = \ln(\cos ax - \sin ax) \implies y(0) = \ln 1 = 0$
$y' = \frac{-a\sin ax - a\cos ax}{\cos ax - \sin ax} \implies y'(0) = -a$
From (i), $y'' + (y')^2 + a^2 = 0 \implies y''(0) + (-a)^2 + a^2 = 0 \implies y''(0) = -2a^2$
Differentiating (i), $y''' + 2y'y'' = 0 \implies y'''(0) + 2(-a)(-2a^2) = 0 \implies y'''(0) = -4a^3$
The Maclaurin series for $y$ is:
$y = -ax - \frac{2a^2}{2!}x^2 - \frac{4a^3}{3!}x^3 + \ldots = -ax - a^2x^2 - \frac{2}{3}a^3x^3 + \ldots$

2. Relate $\cos 2x - \sin 2x$ to $y$ and substitute $a=2$ into the truncated series.

Since $y = \ln(\cos ax - \sin ax)$, taking the exponential gives $\cos ax - \sin ax = e^y$.
For $a=2$, since $x$ is small enough to neglect powers higher than 2, we approximate $y$ to the $x^2$ term:
$y \approx -2x - (2)^2x^2 = -2x - 4x^2$

3. Use the standard Maclaurin series for $e^y$ to expand the expression.

Using the standard expansion $e^y \approx 1 + y + \frac{y^2}{2!}$:
$\cos 2x - \sin 2x = e^y \approx 1 + (-2x - 4x^2) + \frac{1}{2}(-2x - 4x^2)^2$

4. Expand and simplify, discarding powers of $x$ higher than 2, to find $k$.

$\cos 2x - \sin 2x \approx 1 - 2x - 4x^2 + \frac{1}{2}(4x^2)$
$= 1 - 2x - 4x^2 + 2x^2$
$= 1 - 2x - 2x^2$
Comparing this to the given form $1 + kx + kx^2$, we obtain:
$k = -2$

1. Use standard series from MF26 to expand $\cos 2x$ and $\sin 2x$ up to the term in $x^2$.

From the List of Formulae (MF26), the standard series expansions are $\cos x \approx 1 - \frac{x^2}{2!}$ and $\sin x \approx x$ for small $x$.
Replacing $x$ with $2x$, we obtain:
$\cos 2x \approx 1 - \frac{(2x)^2}{2!} = 1 - 2x^2$
$\sin 2x \approx 2x$

2. Subtract the expansions to verify the result from part (iii).

$\cos 2x - \sin 2x \approx (1 - 2x^2) - (2x) = 1 - 2x - 2x^2$
This simplifies to the form $1 + kx + kx^2$ with $k = -2$, identical to the correct answer found in (iii), thus verifying it.