Let C 1 be the curve obtained by the solution of differential equation 2 x y d y d x = y 2 - x 2 ,   x…

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

put $y=vx$

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

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

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

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

put $x=1,y=1,C=ln2$

$ln\left(\frac{y^2}{x^2}+1\right)=-lnx+ln2$

$\Rightarrow x^2+y^2-2x=0$ (Curve $C_1$)

Similarly,

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

Put $y=v x$

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

Required area $=2{\int }_0^1\left(\sqrt{2x-x^2}-x\right)dx=\frac{\pi }{2}-1$ sq. units