$\begin{array}{l}\text{Let }{C}_1\text{\hspace{0.17em} be the curve obtained by the solution of the differential equation }2xy\frac{dy}{dx}=y^2-x^2\\ \text{Let the curve }{C}_2\text{\hspace{0.17em}be the solution of }\frac{2xy}{x^2-y^2}=\frac{dy}{dx}.\\ \text{If both curves pass through the point }\left(1,1\right),\text{then the area enclosed by the curve }{C}_1\text{\hspace{0.17em} and }{C}_2\text{\hspace{0.17em} is equal to}\end{array}$

$\text{ The given differential equation }\frac{dy}{dx}=\frac{y^2-x^2}{2xy},\text{ it is homogeneous equation. }$

$\text{ Put }y=vx$

$\begin{array}{l}v+x\frac{dv}{dx}=\frac{v^2{x}^2-x^2}{2v{x}^2}=\frac{v^2-1}{2v}x\frac{dv}{dx}=\frac{v^2-1-2v^2}{2v}=\frac{-\left(v^2+1\right)}{2v}\\ \Rightarrow \frac{2v}{v^2+1}dv=-\frac{dx}{x}\ln \left(v^2+1\right)=-\ell nx+\ln c\Rightarrow v^2+1=\frac{c}{x}\\ \Rightarrow \frac{y^2}{x^2}+1=\frac{c}{x}\Rightarrow x^2+y^2=cx\text{ If pass through }(1,1) \therefore x^2+y^2-2x=0\end{array}$

Similarly second differential equation is $\frac{dx}{dy}=\frac{x^2-y^2}{2xy}$

Equation of curve is $x^2+y^2-2y=0$

$\text{Area of the region required is equal to the double the ( area of the quarter circle - area of the triangle )}$

$=\left(\frac{1}{4}\times \pi \times 1^2-\frac{1}{2}\times 1\times 1\right)\times 2=\left(\frac{\pi }{2}-1\right)\text{\hspace{0.17em}square units}$