The variables $x$ and $y$ are related by the equation $y = x - \cos^{-1} x$, where $0 \le x < 1$ and $0 < \cos^{-1} x \le \frac{\pi}{2}$.

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

$y = x - \cos^{-1} x$
$\frac{dy}{dx} = 1 - \left(-\frac{1}{\sqrt{1-x^2}}\right)$
$\frac{dy}{dx} = 1 + \frac{1}{\sqrt{1-x^2}}$

2. Rearrange the expression to isolate the algebraic fraction.

$\frac{dy}{dx} - 1 = (1-x^2)^{-\frac{1}{2}}$

3. Differentiate again with respect to $x$ to find the second derivative.

$\frac{d^2y}{dx^2} = -\frac{1}{2}(1-x^2)^{-\frac{3}{2}}(-2x)$
$\frac{d^2y}{dx^2} = x(1-x^2)^{-\frac{3}{2}}$

4. Substitute the first derivative expression back into the second derivative to complete the proof.

$\frac{d^2y}{dx^2} = x\left((1-x^2)^{-\frac{1}{2}}\right)^3$
$\frac{d^2y}{dx^2} = x\left(\frac{dy}{dx} - 1\right)^3 \quad \text{(shown)}$

1. Evaluate $y$ at $x = 0$.

$y = 0 - \cos^{-1}(0)$
Since $\cos\left(\frac{\pi}{2}\right) = 0$ and the principal value is in $\left[0, \pi\right]$, $\cos^{-1}(0) = \frac{\pi}{2}$.
$y = -\frac{\pi}{2}$

2. Differentiate the given equation to find $\frac{dy}{dx}$ and evaluate at $x=0$.

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

3. Use the result from part (i) to evaluate $\frac{d^2y}{dx^2}$ at $x=0$.

From part (i), we have $\frac{d^2 y}{dx^2} = x\left(\frac{dy}{dx}-1\right)^3$.
Substituting $x = 0$ and $\frac{dy}{dx} = 2$:
$\frac{d^2 y}{dx^2} = 0 \times (2 - 1)^3 = 0$

1. Determine the values of $y$, $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ at $x = 0$ (results from part ii).

At $x = 0$,
$y = 0 - \cos^{-1}(0) = -\frac{\pi}{2}$
$\frac{dy}{dx} = 1 - \left(-\frac{1}{\sqrt{1-x^2}}\right) = 1 + \frac{1}{\sqrt{1-x^2}} = 2$
$\frac{d^2 y}{dx^2} = 0 \times (2 - 1)^3 = 0$

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

Given $\frac{d^2 y}{dx^2} = x\left(\frac{dy}{dx}-1\right)^3$, we differentiate both sides with respect to $x$ using the product and chain rules:
$\frac{d^3 y}{dx^3} = 1 \cdot \left(\frac{dy}{dx}-1\right)^3 + x \cdot 3\left(\frac{dy}{dx}-1\right)^2 \frac{d^2 y}{dx^2}$

3. Evaluate $\frac{d^3y}{dx^3}$ at $x=0$.

Substitute $x = 0$, $\frac{dy}{dx} = 2$, and $\frac{d^2y}{dx^2} = 0$:
$\frac{d^3 y}{dx^3} = (2-1)^3 + 0 \cdot 3(2-1)^2(0) = 1^3 = 1$

4. Formulate the Maclaurin series expansion for $y$ up to the term in $x^3$.

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 evaluated values:
$y = -\frac{\pi}{2} + 2x + \frac{0}{2!}x^2 + \frac{1}{3!}x^3$
$y = -\frac{\pi}{2} + 2x + \frac{1}{6}x^3$

1. Concisely evaluate the results from earlier parts to obtain the Maclaurin series for $y$ up to the term in $x^3$.

From part (ii), at $x = 0$:
$y = 0 - \cos^{-1}(0) = -\frac{\pi}{2}$
$\frac{dy}{dx} = 1 - \left(-\frac{1}{\sqrt{1-0^2}}\right) = 2$
$\frac{d^2 y}{dx^2} = 0\cdot(2-1)^3 = 0$

From part (iii), differentiating the result in (i) gives:
$\frac{d^3 y}{dx^3} = 1\cdot\left(\frac{dy}{dx}-1\right)^3 + x\cdot 3\left(\frac{dy}{dx}-1\right)^2 \frac{d^2 y}{dx^2}$
At $x = 0$, $\frac{d^3 y}{dx^3} = (2-1)^3 + 0 = 1$

So the Maclaurin series for $y$ is:
$y = -\frac{\pi}{2} + 2x + \frac{0}{2!}x^2 + \frac{1}{3!}x^3 + \dots$
$y = -\frac{\pi}{2} + 2x + \frac{1}{6}x^3 + \dots$

2. Recognise that the required expression is the first derivative $\frac{dy}{dx}$, and differentiate the series for $y$ to deduce the answer.

Notice that $\frac{dy}{dx} = \frac{d}{dx}\left(x - \cos^{-1} x\right) = 1 - \left(-\frac{1}{\sqrt{1-x^2}}\right) = 1 + \frac{1}{\sqrt{1-x^2}}$.

Therefore, we can deduce the required series by differentiating the series expansion of $y$ with respect to $x$:
$1 + \frac{1}{\sqrt{1-x^2}} \approx \frac{d}{dx} \left(-\frac{\pi}{2} + 2x + \frac{1}{6}x^3\right)$
$1 + \frac{1}{\sqrt{1-x^2}} \approx 2 + \frac{3}{6}x^2$
$= 2 + \frac{1}{2}x^2$