Let y = y(x) be a function of x satisfying y 1 - x ^ 2 = k - x 1 -...

$x = \dfrac{1}{2}, y = \dfrac{- 1}{4} ⇒ xy = \dfrac{- 1}{8}$

$y . \dfrac{\left(- 2x\right)}{2\sqrt{1 - {x}^{2}}} + {y}^{'} . \sqrt{1 - {x}^{2}} = 0 - \left\{1 . \sqrt{1 - {y}^{2}} + \dfrac{x . \left(- 2y\right)}{2\sqrt{1 - {y}^{2}}}{y}^{'}\right\}$

$- \dfrac{xy}{\sqrt{1 - {x}^{2}}} + {y}^{'}\sqrt{1 - {x}^{2}} = - \sqrt{1 - {y}^{2}} + \dfrac{xy . {y}^{'}}{\sqrt{1 - {y}^{2}}}$

${y}^{'}\left(\sqrt{1 - {x}^{2}} - \dfrac{xy}{\sqrt{1 - {y}^{2}}}\right) = \dfrac{xy}{\sqrt{1 - {x}^{2}}} - \sqrt{1 - {y}^{2}}$

${y}^{'}\left(\dfrac{\sqrt{3}}{2} + \dfrac{1}{8 . \dfrac{\sqrt{15}}{4}}\right) = \dfrac{- 1}{8 . \sqrt{\dfrac{3}{4}}} - \dfrac{\sqrt{15}}{4}$

${y}^{'}\left(\dfrac{\sqrt{45} + 1}{2\sqrt{15}}\right) = - \dfrac{\left(1 + \sqrt{45}\right)}{4\sqrt{3}}$

${y}^{'} = - \dfrac{\sqrt{15}}{2\sqrt{3}} = - \dfrac{\sqrt{5}}{2}$