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

1. Rewrite the given logarithmic equation into an exponential form to avoid the quotient rule and simplify implicit differentiation.

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

2. Differentiate both sides implicitly with respect to $x$ for the first time.

$\frac{d}{dx}(e^y) = \frac{d}{dx}(\cos ax - \sin ax)$
$e^y \frac{dy}{dx} = -a\sin ax - a\cos ax$

3. Differentiate a second time using the product rule and substitute the original expression to obtain the required equation.

Applying the product rule on the left hand side:
$e^y \frac{d^2y}{dx^2} + e^y \left(\frac{dy}{dx}\right)^2 = -a^2\cos ax + a^2\sin ax$
$e^y \left[ \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 \right] = -a^2(\cos ax - \sin ax)$
Since $e^y = \cos ax - \sin ax$, we can substitute this into the right hand side:
$e^y \left[ \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 \right] = -a^2 e^y$
Dividing both sides by $e^y$ (since $e^y > 0$ for real $y$):
$\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = -a^2$
$\frac{d^2 y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + a^2 = 0 \quad \text{(shown)}$

1. Evaluate $y$, $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ at $x=0$.

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

Differentiating $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{-a\sin ax - a\cos ax}{\cos ax - \sin ax}$
At $x = 0$:
$\frac{dy}{dx} = \frac{-a(0) - a(1)}{1 - 0} = -a$

Using the result from (i), $\frac{d^2 y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + a^2 = 0$. Substitute $x = 0$:
$\frac{d^2 y}{dx^2} + (-a)^2 + a^2 = 0 \implies \frac{d^2 y}{dx^2} = -2a^2$

2. Differentiate the equation from (i) with respect to $x$ to find $\frac{d^3y}{dx^3}$ at $x=0$.

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

Substitute the known values at $x = 0$:
$\frac{d^3 y}{dx^3} + 2(-a)(-2a^2) = 0$
$\frac{d^3 y}{dx^3} + 4a^3 = 0 \implies \frac{d^3 y}{dx^3} = -4a^3$

3. Substitute the evaluated derivatives into the Maclaurin series formula.

The Maclaurin series is given by:
$y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots$

Substituting the values:
$y = 0 + (-a)x + \frac{-2a^2}{2}x^2 + \frac{-4a^3}{6}x^3$
$y = -ax - a^2 x^2 - \frac{2}{3}a^3 x^3$

1. Find the Maclaurin series for $y$ up to the $x^2$ term (as required from part (ii)).

To use the result from part (ii), we first calculate the Maclaurin series coefficients for $y$ up to $x^2$.
When $x = 0$, $y = \ln(\cos 0 - \sin 0) = \ln 1 = 0$.
$y' = \frac{d}{dx}[\ln(\cos ax - \sin ax)] = \frac{-a\sin ax - a\cos ax}{\cos ax - \sin ax} \implies y'(0) = -a$.
Using the result from (i), $y'' + (y')^2 + a^2 = 0$.
Substituting $x = 0$, $y''(0) + (-a)^2 + a^2 = 0 \implies y''(0) = -2a^2$.
Thus, the Maclaurin series for $y$ up to $x^2$ is:
$y \approx y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 = -ax - a^2 x^2$

2. Substitute $a = 2$ and express $\cos 2x - \sin 2x$ as an exponential function.

Substitute $a = 2$ into the series and neglect powers of $x$ higher than $2$:
$y \approx -2x - 4x^2$
Since $y = \ln(\cos 2x - \sin 2x)$, we have:
$\ln(\cos 2x - \sin 2x) \approx -2x - 4x^2$
$\cos 2x - \sin 2x \approx e^{-2x - 4x^2}$

3. Expand the exponential expression using standard series to find $k$.

Using the standard Maclaurin series expansion $e^z \approx 1 + z + \frac{z^2}{2!}$ for small $z$:
$e^{-2x - 4x^2} \approx 1 + (-2x - 4x^2) + \frac{(-2x - 4x^2)^2}{2}$
Neglecting powers of $x$ higher than $2$:
$e^{-2x - 4x^2} \approx 1 - 2x - 4x^2 + \frac{4x^2}{2}$
$= 1 - 2x - 4x^2 + 2x^2$
$= 1 - 2x - 2x^2$
Comparing this with $1 + kx + kx^2$, we have:
$k = -2$

1. Expand $\cos 2x$ and $\sin 2x$ up to the term in $x^2$ using the standard Maclaurin series from MF26.

From the List of Formulae (MF26):
$\cos x = 1 - \frac{x^2}{2!} + \dots$
$\sin x = x - \frac{x^3}{3!} + \dots$
Substituting $x$ with $2x$ and neglecting powers of $x$ higher than $2$:
$\cos 2x \approx 1 - \frac{(2x)^2}{2!} = 1 - 2x^2$
$\sin 2x \approx 2x$

2. Subtract the expansions to form the required expression and verify the constant $k$.

$\cos 2x - \sin 2x \approx (1 - 2x^2) - (2x)$
$= 1 - 2x - 2x^2$
This explicitly matches the mathematical form $1 + kx + kx^2$ found in part (iii) and verifies that the constant is indeed $k = -2$.