已知 $y=(\cos^{-1} x)^2$。

1. Differentiate $y$ with respect to $x$ using the chain rule.

$\dfrac{dy}{dx} = 2(\cos^{-1} x) \cdot \left( -\dfrac{1}{\sqrt{1-x^2}} \right)$

2. Rearrange the equation to avoid applying the quotient rule in the subsequent differentiation.

$\sqrt{1-x^2}\dfrac{dy}{dx} = -2\cos^{-1} x$

3. Differentiate both sides implicitly with respect to $x$ using the product rule.

$\sqrt{1-x^2}\dfrac{d^2y}{dx^2} + \dfrac{dy}{dx}\left(\dfrac{1}{2\sqrt{1-x^2}} \cdot (-2x)\right) = -2\left(-\dfrac{1}{\sqrt{1-x^2}}\right)$
$\sqrt{1-x^2}\dfrac{d^2y}{dx^2} - \dfrac{x}{\sqrt{1-x^2}}\dfrac{dy}{dx} = \dfrac{2}{\sqrt{1-x^2}}$

4. Simplify the equation to obtain the required form.

Multiply both sides by $\sqrt{1-x^2}$:
$(1-x^2)\dfrac{d^2y}{dx^2} - x\dfrac{dy}{dx} = 2$ (Shown)

1. Find the values of $y$ and $\frac{dy}{dx}$ at $x=0$.

When $x = 0$,
$y = (\cos^{-1} 0)^2 = \left(\frac{\pi}{2}\right)^2 = \frac{\pi^2}{4}$

$\frac{dy}{dx} = 2(\cos^{-1} x) \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right)$
At $x = 0$, $\frac{dy}{dx} = 2\left(\frac{\pi}{2}\right)(-1) = -\pi$

2. Find the value of $\frac{d^2y}{dx^2}$ at $x=0$ using the result from part (i).

From part (i), we have:
$(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = 2$
Substituting $x = 0$,
$(1 - 0)\frac{d^2y}{dx^2} - 0 = 2$
$\frac{d^2y}{dx^2} = 2$

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

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

Substituting $x = 0$,
$(1-0)\frac{d^3y}{dx^3} - 0 - (-\pi) = 0$
$\frac{d^3y}{dx^3} = -\pi$

4. Form the Maclaurin's series up to the term in $x^3$.

The Maclaurin's 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 known values:
$(\cos^{-1} x)^2 = \frac{\pi^2}{4} + (-\pi)x + \frac{2}{2!}x^2 + \frac{-\pi}{3!}x^3 + \dots$
$(\cos^{-1} x)^2 = \frac{\pi^2}{4} - \pi x + x^2 - \frac{\pi}{6}x^3 + \dots$

1. Obtain the Maclaurin series for $(\cos^{-1} x)^2$ up to the term in $x^2$.

To use the result from part (ii), we calculate the required derivatives at $x=0$.
Let $y = (\cos^{-1} x)^2$.
When $x=0$, $y = (\cos^{-1} 0)^2 = \left(\frac{\pi}{2}\right)^2 = \frac{\pi^2}{4}$.
$\frac{dy}{dx} = 2(\cos^{-1} x)\left(-\frac{1}{\sqrt{1-x^2}}\right) \implies \left.\frac{dy}{dx}\right|_{x=0} = 2\left(\frac{\pi}{2}\right)(-1) = -\pi$.
From $(1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}=2$ (proven in part i), substituting $x=0$ gives $(1)\frac{d^2y}{dx^2} - 0 = 2 \implies \left.\frac{d^2y}{dx^2}\right|_{x=0} = 2$.
Thus, the Maclaurin series up to $x^2$ is:
$(\cos^{-1} x)^2 \approx \frac{\pi^2}{4} - \pi x + \frac{2}{2!}x^2 = \frac{\pi^2}{4} - \pi x + x^2$.

2. Substitute the series into the exponential function and isolate the constant factor $e^{\pi^2/4}$.

Substitute the quadratic approximation into the exponent:
$e^{(\cos^{-1} x)^2} \approx e^{\frac{\pi^2}{4} - \pi x + x^2}$
Using the laws of indices:
$e^{\frac{\pi^2}{4} - \pi x + x^2} = e^{\frac{\pi^2}{4}} e^{-\pi x + x^2}$

3. Expand $e^{-\pi x + x^2}$ using the standard Maclaurin series for $e^u$, neglecting powers of $x$ above $x^2$.

Using the standard series expansion $e^u \approx 1 + u + \frac{u^2}{2!}$ for small $u$:
$e^{-\pi x + x^2} \approx 1 + (-\pi x + x^2) + \frac{(-\pi x + x^2)^2}{2}$
Expanding the squared term and neglecting powers of $x$ above $x^2$ (since $(-\pi x)^2 = \pi^2 x^2$, while cross terms and $(x^2)^2$ produce higher powers):
$e^{-\pi x + x^2} \approx 1 - \pi x + x^2 + \frac{1}{2}(\pi^2 x^2)$
$= 1 - \pi x + \left(1 + \frac{\pi^2}{2}\right)x^2$

4. Equate coefficients to find $a$ and $b$.

Substitute this back into the original expression:
$e^{(\cos^{-1} x)^2} \approx e^{\frac{\pi^2}{4}} \left[ 1 - \pi x + \left(1 + \frac{\pi^2}{2}\right)x^2 \right]$
Comparing this with the given form $e^{\pi^2/4}(1+ax+bx^2)$, we get:
$a = -\pi$ and $b = 1 + \frac{\pi^2}{2}$.