Let a solution $y = y(x)$ of the differential equation $x\sqrt{x^{2} - 1}\,dy - y\sqrt{y^{2} - 1}\,dx = 0$ satisfy $y(2) = \frac{2}{\sqrt{3}}$. STATEMENT-1: $y(x) = \sec\left(\sec^{-1} x - \frac{\pi}{6}\right)$ and STATEMENT-2: $y(x)$ is given by $\frac{1}{y} = \frac{2\sqrt{3}}{x} - \sqrt{1 - \frac{1}{x^{2}}}$

$x\sqrt{x^2 - 1}\,dy - y\sqrt{y^2 - 1}\,dx = 0$

$y(2) = \dfrac{2}{\sqrt3}$

$\text{To find: which option is correct}$

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

$\int \dfrac{dt}{t\sqrt{t^2-1}} = \sec^{-1}t$

$\sec^{-1}y = \sec^{-1}x + C$

$\sec^{-1}2 = \dfrac{\pi}{3},\qquad \sec^{-1}\!\left(\dfrac{2}{\sqrt3}\right) = \dfrac{\pi}{6}$

$\dfrac{\pi}{6} = \dfrac{\pi}{3} + C \Rightarrow C = -\dfrac{\pi}{6}$

$y = \sec\!\left(\sec^{-1}x - \dfrac{\pi}{6}\right)$

$\Rightarrow\ \text{STATEMENT-1 is TRUE}$

$\dfrac{1}{y} = \cos\!\left(\sec^{-1}x - \dfrac{\pi}{6}\right) = \cos A\cos B + \sin A\sin B$

$A = \sec^{-1}x:\ \cos A = \dfrac1x,\ \sin A = \sqrt{1-\dfrac{1}{x^2}},\quad \cos B = \dfrac{\sqrt3}{2},\ \sin B = \dfrac12$

$\dfrac{1}{y} = \dfrac{\sqrt3}{2x} + \dfrac12\sqrt{1-\dfrac{1}{x^2}} \ \ne\ \dfrac{2\sqrt3}{x} - \sqrt{1-\dfrac{1}{x^2}}$

$\Rightarrow\ \text{STATEMENT-2 is FALSE}$

$\boxed{\text{option } (3)}$