Show that for $x>1$,

1. Express the given function with a negative rational index and apply the chain rule.

$\begin{aligned}\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{1}{\sqrt{x^2-1}}\right) &= \dfrac{\mathrm{d}}{\mathrm{d}x}\left((x^2-1)^{-\frac{1}{2}}\right) \\ &= -\frac{1}{2}(x^2-1)^{-\frac{3}{2}} \cdot \dfrac{\mathrm{d}}{\mathrm{d}x}(x^2-1) \end{aligned}$

2. Evaluate the inner derivative and simplify the expression to the required form.

$\begin{aligned} -\frac{1}{2}(x^2-1)^{-\frac{3}{2}} \cdot (2x) &= -x(x^2-1)^{-\frac{3}{2}} \\ &= -\dfrac{x}{(x^2-1)^{\frac{3}{2}}} \end{aligned}$

1. Apply the chain rule using the standard derivative for the inverse sine function.

$\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\sin^{-1}\dfrac{1}{x}\right) = \dfrac{1}{\sqrt{1 - \left(\frac{1}{x}\right)^2}} \cdot \dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^{-1}\right)$

2. Simplify the expression algebraically, taking careful note of the condition $x > 1$ when extracting the square root.

$= \dfrac{1}{\sqrt{\dfrac{x^2 - 1}{x^2}}} \cdot \left(-x^{-2}\right)$
$= \dfrac{1}{\dfrac{\sqrt{x^2 - 1}}{|x|}} \cdot \left(-\dfrac{1}{x^2}\right)$
Since it is given that $x > 1$, we have $|x| = x$. Thus:
$= \dfrac{x}{\sqrt{x^2 - 1}} \cdot \left(-\dfrac{1}{x^2}\right)$
$= -\dfrac{1}{x\sqrt{x^2-1}} \quad \text{(shown)}$

1. Find the limits of the parameter $t$ corresponding to the boundary lines $x = \frac{1}{\sqrt{3}}$ and $x = \sqrt{3}$.

When $x = \frac{1}{\sqrt{3}}$,
$\frac{1}{\sqrt{t^2-1}} = \frac{1}{\sqrt{3}}$
$t^2 - 1 = 3 \implies t^2 = 4 \implies t = 2 \quad (\text{since } t > 1)$

When $x = \sqrt{3}$,
$\frac{1}{\sqrt{t^2-1}} = \sqrt{3}$
$t^2 - 1 = \frac{1}{3} \implies t^2 = \frac{4}{3} \implies t = \frac{2}{\sqrt{3}} \quad (\text{since } t > 1)$

2. Determine $\frac{\mathrm{d}x}{\mathrm{d}t}$ using the result from part (i).

From part (i), we are given $\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{\sqrt{x^2-1}}\right)=-\frac{x}{(x^2-1)^{3/2}}$.
Replacing $x$ with $t$, we obtain:
$\frac{\mathrm{d}x}{\mathrm{d}t} = -\frac{t}{(t^2-1)^{3/2}}$

3. Set up the definite integral for the area and reverse the limits to match the required form.

$\text{Area} = \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} y \,\mathrm{d}x$
Substituting $y = \ln t$, $\mathrm{d}x = \frac{\mathrm{d}x}{\mathrm{d}t}\mathrm{d}t$ and the calculated $t$ limits:
$= \int_{2}^{\frac{2}{\sqrt{3}}} (\ln t) \left( -\frac{t}{(t^2-1)^{3/2}} \right) \,\mathrm{d}t$
Reversing the limits of integration changes the sign of the integral:
$= \int_{\frac{2}{\sqrt{3}}}^{2} \frac{t}{(t^2-1)^{3/2}}(\ln t) \,\mathrm{d}t \quad \text{(Shown)}$

1. Set up the components for integration by parts.

Let $u = \ln t \implies \dfrac{\mathrm{d}u}{\mathrm{d}t} = \dfrac{1}{t}$

Let $\dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{t}{(t^2-1)^{3/2}}$

From the result in (i), $v = -\dfrac{1}{\sqrt{t^2-1}}$

2. Apply the integration by parts formula.

Using the formula $\int u \dfrac{\mathrm{d}v}{\mathrm{d}t} \,\mathrm{d}t = uv - \int v \dfrac{\mathrm{d}u}{\mathrm{d}t} \,\mathrm{d}t$:

$\text{Area} = \left[ -\dfrac{\ln t}{\sqrt{t^2-1}} \right]_{\frac{2}{\sqrt{3}}}^{2} - \int_{\frac{2}{\sqrt{3}}}^{2} \left( -\dfrac{1}{\sqrt{t^2-1}} \right) \left( \dfrac{1}{t} \right) \,\mathrm{d}t$

$\text{Area} = \left[ -\dfrac{\ln t}{\sqrt{t^2-1}} \right]_{\frac{2}{\sqrt{3}}}^{2} + \int_{\frac{2}{\sqrt{3}}}^{2} \dfrac{1}{t\sqrt{t^2-1}} \,\mathrm{d}t$

3. Evaluate the remaining integral using the result from (ii).

From the result in (ii), we know that $\int \dfrac{1}{t\sqrt{t^2-1}} \,\mathrm{d}t = -\sin^{-1}\left(\dfrac{1}{t}\right)$

$\text{Area} = \left[ -\dfrac{\ln t}{\sqrt{t^2-1}} - \sin^{-1} \left( \dfrac{1}{t} \right) \right]_{\frac{2}{\sqrt{3}}}^{2}$

4. Substitute the upper limit of integration.

Substitute $t=2$ into the expression:

$-\dfrac{\ln 2}{\sqrt{2^2-1}} - \sin^{-1} \left( \dfrac{1}{2} \right) = -\dfrac{\ln 2}{\sqrt{3}} - \dfrac{\pi}{6}$

5. Substitute the lower limit of integration.

Substitute $t=\dfrac{2}{\sqrt{3}}$ into the expression:

$-\dfrac{\ln \left( \frac{2}{\sqrt{3}} \right)}{\sqrt{\frac{4}{3}-1}} - \sin^{-1} \left( \dfrac{\sqrt{3}}{2} \right)$

$= -\dfrac{\ln 2 - \frac{1}{2}\ln 3}{\frac{1}{\sqrt{3}}} - \dfrac{\pi}{3}$

$= -\sqrt{3}\ln 2 + \dfrac{\sqrt{3}}{2}\ln 3 - \dfrac{\pi}{3}$

6. Calculate the exact final area by finding the difference.

$\text{Area} = \left( -\dfrac{\sqrt{3}}{3}\ln 2 - \dfrac{\pi}{6} \right) - \left( -\sqrt{3}\ln 2 + \dfrac{\sqrt{3}}{2}\ln 3 - \dfrac{\pi}{3} \right)$

$= -\dfrac{\sqrt{3}}{3}\ln 2 + \sqrt{3}\ln 2 - \dfrac{\sqrt{3}}{2}\ln 3 - \dfrac{\pi}{6} + \dfrac{\pi}{3}$

$= \dfrac{2\sqrt{3}}{3}\ln 2 - \dfrac{\sqrt{3}}{2}\ln 3 + \dfrac{\pi}{6}$