It is given that $y = e^{\cos^{-1} x}$.

1. Find the first derivative $\dfrac{dy}{dx}$ and express it in terms of $y$.

$y = e^{\cos^{-1} x}$

Differentiating with respect to $x$ using the chain rule:
$\frac{dy}{dx} = e^{\cos^{-1} x} \cdot \left( -\frac{1}{\sqrt{1 - x^2}} \right)$

Substitute $y = e^{\cos^{-1} x}$ back into the equation:
$\frac{dy}{dx} = -\frac{y}{\sqrt{1 - x^2}}$

2. Rearrange the equation and square both sides to eliminate the fraction and square root.

Multiply by $\sqrt{1 - x^2}$:
$\sqrt{1 - x^2} \frac{dy}{dx} = -y$

Squaring both sides gives:
$(1 - x^2) \left(\frac{dy}{dx}\right)^2 = y^2$

3. Differentiate implicitly with respect to $x$ using the product and chain rules, then simplify.

Differentiating both sides with respect to $x$:
$(1 - x^2) \frac{d}{dx}\left[\left(\frac{dy}{dx}\right)^2\right] + \left(\frac{dy}{dx}\right)^2 \frac{d}{dx}(1 - x^2) = \frac{d}{dx}(y^2)$

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

Dividing both sides by the common factor $2\frac{dy}{dx}$:
$(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = y \quad \text{(Shown)}$

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

Given $y = e^{\cos^{-1} x}$, when $x = 0$:
$y(0) = e^{\cos^{-1} 0} = e^{\frac{\pi}{2}}$

Differentiating with respect to $x$:
$y' = e^{\cos^{-1} x} \left( -\frac{1}{\sqrt{1 - x^2}} \right)$
When $x = 0$, $y'(0) = e^{\frac{\pi}{2}} (-1) = -e^{\frac{\pi}{2}}$.

From part (i), we have the relation:
$(1 - x^2)y'' - xy' = y$
Substituting $x = 0$, $y(0) = e^{\frac{\pi}{2}}$ and $y'(0) = -e^{\frac{\pi}{2}}$:
$(1 - 0)y''(0) - 0 = e^{\frac{\pi}{2}} \implies y''(0) = e^{\frac{\pi}{2}}$

2. Differentiate the given differential equation to find an expression for $\frac{d^3 y}{dx^3}$.

Differentiating $(1 - x^2)y'' - xy' = y$ with respect to $x$ using the product rule:
$(1 - x^2)y''' + y''(-2x) - (x y'' + y') = y'$

Simplifying the equation:
$(1 - x^2)y''' - 3xy'' - y' = y'$
$(1 - x^2)y''' - 3xy'' - 2y' = 0$

3. Evaluate $\frac{d^3 y}{dx^3}$ at $x=0$ and construct the Maclaurin's series.

Substitute $x = 0$, $y'(0) = -e^{\frac{\pi}{2}}$, and $y''(0) = e^{\frac{\pi}{2}}$ into the differentiated equation:
$(1 - 0)y'''(0) - 0 - 2\left(-e^{\frac{\pi}{2}}\right) = 0$
$y'''(0) = -2e^{\frac{\pi}{2}}$

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 exact values:
$y = e^{\frac{\pi}{2}} + \left(-e^{\frac{\pi}{2}}\right)x + \frac{e^{\frac{\pi}{2}}}{2}x^2 + \frac{-2e^{\frac{\pi}{2}}}{6}x^3 + \dots$
$y = e^{\frac{\pi}{2}} - e^{\frac{\pi}{2}} x + \frac{e^{\frac{\pi}{2}}}{2} x^2 - \frac{e^{\frac{\pi}{2}}}{3} x^3 + \dots$

1. Concise derivation of the result from part (ii) (Maclaurin series for $y$ up to $x^3$).

From part (ii), we first need the Maclaurin series of $y = e^{\cos^{-1} x}$ up to the term in $x^3$.
At $x=0$, $y(0) = e^{\pi/2}$.
$\dfrac{dy}{dx} = -\dfrac{e^{\cos^{-1} x}}{\sqrt{1 - x^2}} \implies y'(0) = -e^{\pi/2}$.
Using $\left(1 - x^2\right)y'' - xy' = y$ from (i):
$y''(0) - 0 = y(0) \implies y''(0) = e^{\pi/2}$.
Repeated differentiation implicitly yields:
$-2xy'' + (1-x^2)y''' - y' - xy'' = y' \implies (1-x^2)y''' - 3xy'' - 2y' = 0$
$y'''(0) - 0 - 2(-e^{\pi/2}) = 0 \implies y'''(0) = -2e^{\pi/2}$.
Thus, the series up to $x^3$ is:
$y \approx e^{\pi/2} - e^{\pi/2}x + \dfrac{e^{\pi/2}}{2}x^2 - \dfrac{2e^{\pi/2}}{6}x^3 = e^{\pi/2}\left(1 - x + \dfrac{x^2}{2} - \dfrac{x^3}{3}\right)$.

2. Recognise the given expression as the first derivative of $y$ and differentiate the series to find its expansion.

Notice that $-\dfrac{e^{\cos^{-1} x}}{\sqrt{1 - x^2}} = \dfrac{dy}{dx}$.
Using the "Hence" instruction, we differentiate the Maclaurin series of $y$ term-by-term to find the expansion of its derivative:
$\dfrac{dy}{dx} \approx \dfrac{d}{dx} \left( e^{\pi/2} - e^{\pi/2}x + \dfrac{e^{\pi/2}}{2}x^2 - \dfrac{e^{\pi/2}}{3}x^3 \right)$
$\dfrac{dy}{dx} \approx -e^{\pi/2} + e^{\pi/2}x - e^{\pi/2}x^2$

3. Use the expansion to estimate the gradient of the tangent at $x = 0.5$.

The gradient of the tangent to the curve is given by $\dfrac{dy}{dx}$.
Substituting $x = 0.5$ into the truncated expansion:
$\text{Gradient} \approx -e^{\pi/2} + e^{\pi/2}(0.5) - e^{\pi/2}(0.5)^2$
$= e^{\pi/2} \left(-1 + \dfrac{1}{2} - \dfrac{1}{4}\right)$
$= -\dfrac{3}{4}e^{\pi/2}$
$\approx -3.61 \text{ (to 3 s.f.)}$