If ${\sin }^{-1}\left(\frac{{\mathrm{x}}^2-{\mathrm{y}}^2}{{\mathrm{x}}^2+{\mathrm{y}}^2}\right)=\log \mathrm{a} \mathrm{then} \frac{\mathrm{dy}}{\mathrm{dx}}$is

Given ${\sin }^{-1}\left(\frac{{\mathrm{x}}^2-{\mathrm{y}}^2}{{\mathrm{x}}^2+{\mathrm{y}}^2}\right)=\log \mathrm{a}$

Apply sine function on both sides

$\frac{{\mathrm{x}}^2-{\mathrm{y}}^2}{{\mathrm{x}}^2+{\mathrm{y}}^2}=\sin (\log \mathrm{a})$

Apply Componendo and Dividendo rule

$\begin{array}{rcl}\frac{2{\mathrm{x}}^2}{-2{\mathrm{y}}^2}& =& \frac{\sin (\log \mathrm{a})+1}{\sin (\log \mathrm{a})-1}\\ \frac{x^2}{y^2}& =& \frac{1+\sin \left(\log a\right)}{1-\sin \left(\log a\right)}\end{array}$

Observing that the right hand side of the above equation is constant

Hence

$\frac{y^2}{x^2}=k$

Differentiate both sides with respect to $x$

$\begin{array}{rcl}\frac{x^2\left(2y{\displaystyle \frac{dy}{dx}}\right)-2x\left(y^2\right)}{x^4}& =& 0\\ x\frac{dy}{dx}-y& =& 0\\ \frac{dy}{dx}& =& \boxed{\frac{\mathbf{y}}{\mathbf{x}}}\end{array}$