Let y=y(x) be a function of x satisfying  $y\sqrt{1-x^2}=k-x\sqrt{1-y^2}$where k is a constant $y\left(\frac{1}{2}\right)=-\frac{1}{4}$and . Then  $\frac{dy}{dx}$at $x=\frac{1}{2}$ is equal to:

At $x=\frac{1}{2},y=\frac{-1}{4}\Rightarrow xy=\frac{-1}{8}$.

Differentiating w.r.t. x, we get

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

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

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

$\Rightarrow y\text{'}\left(\frac{\sqrt{3}}{2}+\frac{1}{8.\frac{\sqrt{15}}{4}}\right)=\frac{-1}{8{\displaystyle \frac{\sqrt{3}}{2}}}-\frac{\sqrt{15}}{4}$

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

$\Rightarrow y\text{'}={\overline{)\frac{dy}{dx}}}_{x=\frac{1}{2}}=-\frac{\sqrt{5}}{2}$