Let C _ 1 be the curve obtained by the solution of differential...

$\dfrac{dy}{dx} = \dfrac{{y}^{2} - {x}^{2}}{2xy}, x \in \left(0, \infty\right)$

put $y = vx$

$x\dfrac{dv}{dx} + v = \dfrac{{v}^{2} - 1}{2v}$

$\dfrac{2v}{{v}^{2} + 1}dv = - \dfrac{dx}{x}$

Integrate,
$\ln \left({v}^{2} + 1\right) = - \ln x + C$

$\ln \left(\dfrac{{y}^{2}}{{x}^{2}} + 1\right) = - \ln x + C$

put $x = 1, y = 1, C = \ln 2$

$\ln \left(\dfrac{{y}^{2}}{{x}^{2}} + 1\right) = - \ln x + \ln 2$

$⇒ {x}^{2} + {y}^{2} - 2x = 0$ (Curve ${C}_{1}$)

Similarly,

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

Put $y = vx$

${x}^{2} + {y}^{2} - 2y = 0$

Required area $= 2{∫}_{0}^{1}\left(\sqrt{2x - {x}^{2}} - x\right)dx = \dfrac{\pi}{2} - 1$ sq. units