If =k \) (a constant) then

Given: \(\cos^{-1}\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right)=k\). Let \(c = \cos k\), so \(\frac{x^{2}-y^{2}}{x^{2}+y^{2}} = c\). Differentiating implicitly with respect to \(x\): \(-\frac{1}{\sqrt{1-\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right)^2}} \cdot \frac{d}{dx}\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right) = 0\), hence \(\frac{d}{dx}\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right) = 0\). Compute derivative: \(\frac{(2x-2yy')(x^{2}+y^{2}) - (x^{2}-y^{2})(2x+2yy')}{(x^{2}+y^{2})^{2}} = 0\). Simplify numerator: \(2(x-yy')(x^{2}+y^{2}) - 2(x^{2}-y^{2})(x+yy') = 0\). Divide by 2, expand: \((x-yy')(x^{2}+y^{2}) = (x^{2}-y^{2})(x+yy')\). Expand: \(x^{3}+xy^{2}-x^{2}yy'-y^{3}y' = x^{3}+x^{2}yy'-xy^{2}-y^{3}y'\). Cancel \(x^{3}\) and \(-y^{3}y'\), get \(xy^{2}-x^{2}yy' = x^{2}yy'-xy^{2}\). Rearranging: \(2xy^{2} = 2x^{2}yy'\), so \(y = xy'\), thus \(\frac{dy}{dx} = \frac{y}{x}\).