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

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

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

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

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

2. Rearrange and square both sides to eliminate the square root, making the next differentiation simpler.

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

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

3. Differentiate implicitly with respect to $x$ and simplify to obtain the required expression.

Differentiating both sides with respect to $x$ using the product and chain rules:

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

Dividing throughout by $2\frac{\mathrm{d}y}{\mathrm{d}x}$ (since $\frac{\mathrm{d}y}{\mathrm{d}x} \neq 0$):

$(1 - x^2)\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} - x\frac{\mathrm{d}y}{\mathrm{d}x} = y$ (Shown)

1. Evaluate $y$ and the first derivative $\frac{\mathrm{d}y}{\mathrm{d}x}$ at $x=0$.

When $x = 0$, $y = \mathrm{e}^{\cos^{-1}(0)} = \mathrm{e}^{\frac{\pi}{2}}$.

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

When $x = 0$,
$\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{\frac{\pi}{2}} \cdot \left(\frac{-1}{\sqrt{1-0}}\right) = -\mathrm{e}^{\frac{\pi}{2}}$

2. Use the differential equation from part (i) to evaluate the second derivative at $x=0$.

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

Substituting $x = 0$, $y = \mathrm{e}^{\frac{\pi}{2}}$, and $\frac{\mathrm{d}y}{\mathrm{d}x} = -\mathrm{e}^{\frac{\pi}{2}}$:
$(1 - 0)\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} - 0 = \mathrm{e}^{\frac{\pi}{2}}$
$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \mathrm{e}^{\frac{\pi}{2}}$

3. Differentiate the identity from part (i) implicitly with respect to $x$ to evaluate the third derivative at $x=0$.

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

Substituting $x = 0$ and $\frac{\mathrm{d}y}{\mathrm{d}x} = -\mathrm{e}^{\frac{\pi}{2}}$:
$(1 - 0)\frac{\mathrm{d}^3 y}{\mathrm{d}x^3} - 0 - 2\left(-\mathrm{e}^{\frac{\pi}{2}}\right) = 0$
$\frac{\mathrm{d}^3 y}{\mathrm{d}x^3} = -2\mathrm{e}^{\frac{\pi}{2}}$

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

The Maclaurin's series is given by:
$y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots$
$y = \mathrm{e}^{\frac{\pi}{2}} + \left(-\mathrm{e}^{\frac{\pi}{2}}\right)x + \frac{\mathrm{e}^{\frac{\pi}{2}}}{2}x^2 + \frac{-2\mathrm{e}^{\frac{\pi}{2}}}{6}x^3 + \dots$
$y = \mathrm{e}^{\frac{\pi}{2}} - \mathrm{e}^{\frac{\pi}{2}}x + \frac{1}{2}\mathrm{e}^{\frac{\pi}{2}}x^2 - \frac{1}{3}\mathrm{e}^{\frac{\pi}{2}}x^3 + \dots$

1. Derive the Maclaurin series for $y$ up to $x^3$ as required from part (ii).

From part (ii):
At $x=0$, $y(0) = \mathrm{e}^{\cos^{-1} 0} = \mathrm{e}^{\frac{\pi}{2}}$.
$\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{\mathrm{e}^{\cos^{-1} x}}{\sqrt{1 - x^2}} \implies y'(0) = -\mathrm{e}^{\frac{\pi}{2}}$.
Using the given equation $(1 - x^2)y'' - xy' = y$, substituting $x=0$ yields $y''(0) = \mathrm{e}^{\frac{\pi}{2}}$.
Differentiating the given equation implicitly yields $(1-x^2)y''' - 3xy'' - 2y' = 0$, giving $y'''(0) = -2\mathrm{e}^{\frac{\pi}{2}}$.
Thus, the Maclaurin series is $y \approx \mathrm{e}^{\frac{\pi}{2}} \left( 1 - x + \frac{1}{2}x^2 - \frac{1}{3}x^3 \right)$.

2. Relate the target expression to $\frac{\mathrm{d}y}{\mathrm{d}x}$ and find its expansion by differentiating the series.

Observe that the required expression $\frac{\mathrm{e}^{\cos^{-1} x}}{\sqrt{1 - x^2}}$ is exactly $-\frac{\mathrm{d}y}{\mathrm{d}x}$.
Differentiating the Maclaurin series for $y$ term-by-term:
$\frac{\mathrm{d}y}{\mathrm{d}x} \approx \mathrm{e}^{\frac{\pi}{2}} \left( -1 + x - x^2 \right)$
Multiplying by $-1$ gives the required expansion up to $x^2$:
$\frac{\mathrm{e}^{\cos^{-1} x}}{\sqrt{1 - x^2}} \approx \mathrm{e}^{\frac{\pi}{2}} \left( 1 - x + x^2 \right)$

3. Estimate the gradient of the tangent at $x = 0.5$.

The gradient of the tangent to the curve is given by $\frac{\mathrm{d}y}{\mathrm{d}x}$.
Using the series for $\frac{\mathrm{d}y}{\mathrm{d}x}$ derived above:
$\text{Gradient} \approx \mathrm{e}^{\frac{\pi}{2}} \left( -1 + x - x^2 \right)$
Substitute $x = 0.5$:
$\text{Gradient} \approx \mathrm{e}^{\frac{\pi}{2}} \left( -1 + 0.5 - (0.5)^2 \right)$
$\text{Gradient} \approx \mathrm{e}^{\frac{\pi}{2}} (-0.75) = -\frac{3}{4}\mathrm{e}^{\frac{\pi}{2}}$