已知 $y = \sin^{-1} x \cos^{-1} x$，其中 $-1 \le x \le 1$。

1. Differentiate $y$ with respect to $x$ using the product rule and standard derivatives of inverse trigonometric functions.

$y = \sin^{-1} x \cos^{-1} x$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \left( \frac{1}{\sqrt{1-x^2}} \right) \cos^{-1} x + \sin^{-1} x \left( \frac{-1}{\sqrt{1-x^2}} \right)$

2. Simplify the expression and rearrange it to match the required form.

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\cos^{-1} x - \sin^{-1} x}{\sqrt{1-x^2}}$
$\sqrt{1-x^2} \frac{\mathrm{d}y}{\mathrm{d}x} = \cos^{-1} x - \sin^{-1} x$

1. Differentiate both sides with respect to $x$, applying the product rule on the left hand side.

Differentiating $\sqrt{1-x^2} \frac{\mathrm{d}y}{\mathrm{d}x} = \cos^{-1} x - \sin^{-1} x$ with respect to $x$:
$\sqrt{1-x^2} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \frac{\mathrm{d}y}{\mathrm{d}x} \left( \frac{1}{2\sqrt{1-x^2}} \right)(-2x) = \frac{\mathrm{d}}{\mathrm{d}x}(\cos^{-1} x - \sin^{-1} x)$
$\sqrt{1-x^2} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} - \frac{x}{\sqrt{1-x^2}} \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x}(\cos^{-1} x - \sin^{-1} x)$

2. Differentiate the inverse trigonometric functions on the right hand side, then simplify the equation to the required form.

Since $\frac{\mathrm{d}}{\mathrm{d}x}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}$ and $\frac{\mathrm{d}}{\mathrm{d}x}(\cos^{-1} x) = -\frac{1}{\sqrt{1-x^2}}$:
$\sqrt{1-x^2} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} - \frac{x}{\sqrt{1-x^2}} \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x^2}}$
$\sqrt{1-x^2} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} - \frac{x}{\sqrt{1-x^2}} \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{2}{\sqrt{1-x^2}}$
Multiplying both sides by $\sqrt{1-x^2}$:
$(1-x^2) \frac{\mathrm{d}^2y}{\mathrm{d}x^2} - x \frac{\mathrm{d}y}{\mathrm{d}x} = -2 \quad \text{(shown)}$

1. Evaluate the first derivative at $x = 0$ to find $A$.

From (i), $\sqrt{1-x^2} \frac{\mathrm{d}y}{\mathrm{d}x} = \cos^{-1} x - \sin^{-1} x$.
Substituting $x = 0$:
$\sqrt{1-0} \left. \frac{\mathrm{d}y}{\mathrm{d}x} \right|_{x=0} = \cos^{-1}(0) - \sin^{-1}(0)$
$y'(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$
By definition of Maclaurin series, $A = y'(0) = \frac{\pi}{2}$.

2. Evaluate the second derivative at $x = 0$ to find $B$.

From (ii), $(1-x^2) \frac{\mathrm{d}^2y}{\mathrm{d}x^2} - x \frac{\mathrm{d}y}{\mathrm{d}x} = -2$.
Substituting $x = 0$:
$(1-0)y''(0) - 0 = -2$
$y''(0) = -2$
By definition of Maclaurin series, $B = \frac{y''(0)}{2!} = \frac{-2}{2} = -1$.

3. Differentiate the equation in (ii) implicitly to find an expression involving the third derivative.

Differentiating $(1-x^2) \frac{\mathrm{d}^2y}{\mathrm{d}x^2} - x \frac{\mathrm{d}y}{\mathrm{d}x} = -2$ with respect to $x$:
$(1-x^2)\frac{\mathrm{d}^3y}{\mathrm{d}x^3} + (-2x)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - \left(x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \frac{\mathrm{d}y}{\mathrm{d}x}\right) = 0$
$(1-x^2)\frac{\mathrm{d}^3y}{\mathrm{d}x^3} - 3x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - \frac{\mathrm{d}y}{\mathrm{d}x} = 0$

4. Evaluate the third derivative at $x = 0$ to find $C$.

Substituting $x = 0$ into the differentiated equation:
$(1-0)y'''(0) - 0 - y'(0) = 0$
$y'''(0) = y'(0) = \frac{\pi}{2}$
By definition of Maclaurin series, $C = \frac{y'''(0)}{3!} = \frac{\pi/2}{6} = \frac{\pi}{12}$.

1. Concisely calculate the constants $A$, $B$, and $C$ from part (iii).

At $x=0$, $y(0) = 0$.
From (i), $y'(0) = \frac{\pi}{2}$.
From (ii), $y''(0) = -2$.
Differentiating (ii): $(1-x^2)y''' - 3xy'' - y' = 0$.
At $x=0$, $y'''(0) - y'(0) = 0 \implies y'''(0) = \frac{\pi}{2}$.
The Maclaurin series is $y \approx y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 = \frac{\pi}{2}x - x^2 + \frac{\pi}{12}x^3$.
Thus, $A = \frac{\pi}{2}$, $B = -1$, $C = \frac{\pi}{12}$.

2. Calculate the student's estimated value for $x = 0.8$.

$\text{Estimate} = \frac{\pi}{2}(0.8) - (0.8)^2 + \frac{\pi}{12}(0.8)^3$
$= 0.4\pi - 0.64 + \frac{16\pi}{375} \approx 0.751$

3. Calculate the exact value of the expression at $x = 0.8$.

$\text{Exact value} = \sin^{-1}(0.8) \cos^{-1}(0.8)$
$\approx (0.927295)(0.643501) \approx 0.597$

4. Compare the exact and estimated values to conclude if the estimate is good.

The exact value ($0.597$) differs significantly from the estimated value ($0.751$). Thus, the estimate is not a good one. This is expected because the Maclaurin series approximation is most accurate near $x = 0$, and $0.8$ is not sufficiently small.