Given relation

$y\sqrt{1-x^2}=k-x\sqrt{1-y^2}$

Rearranging, $y\sqrt{1-x^2}+x\sqrt{1-y^2}=k$

Since $k$ is constant, differentiate implicitly with respect to $x$.

Differentiate both sides

Differentiate $y\sqrt{1-x^2}+x\sqrt{1-y^2}=k$

For the first term: $\frac{d}{dx}\left(y\sqrt{1-x^2}\right)=\frac{dy}{dx}\sqrt{1-x^2}+y\cdot\frac{-x}{\sqrt{1-x^2}}$

For the second term: $\frac{d}{dx}\left(x\sqrt{1-y^2}\right)=\sqrt{1-y^2}+x\cdot\frac{1}{2}(1-y^2)^{-1/2}(-2y)\frac{dy}{dx}$ $=\sqrt{1-y^2}-\frac{xy}{\sqrt{1-y^2}}\frac{dy}{dx}$

Hence, $\frac{dy}{dx}\sqrt{1-x^2}-\frac{xy}{\sqrt{1-x^2}}+\sqrt{1-y^2}-\frac{xy}{\sqrt{1-y^2}}\frac{dy}{dx}=0$

Collecting $\dfrac{dy}{dx}$ terms, $\frac{dy}{dx}\left(\sqrt{1-x^2}-\frac{xy}{\sqrt{1-y^2}}\right)=\frac{xy}{\sqrt{1-x^2}}-\sqrt{1-y^2}$

Therefore, $\frac{dy}{dx}=\frac{\dfrac{xy}{\sqrt{1-x^2}}-\sqrt{1-y^2}}{\sqrt{1-x^2}-\dfrac{xy}{\sqrt{1-y^2}}}$

Substitute the point

Given: $x=\frac12,\qquad y=-\frac14$

Compute: $\sqrt{1-x^2}=\sqrt{1-\frac14}=\sqrt{\frac34}=\frac{\sqrt3}{2}$

$\sqrt{1-y^2}=\sqrt{1-\frac{1}{16}}=\sqrt{\frac{15}{16}}=\frac{\sqrt{15}}{4}$

Also, $xy=\frac12\cdot\left(-\frac14\right)=-\frac18$

Evaluate numerator

Evaluate denominator

Simplify the ratio

$\frac{dy}{dx}=\frac{-\frac{1}{4\sqrt3}-\frac{\sqrt{15}}{4}}{\frac{\sqrt3}{2}+\frac{1}{2\sqrt{15}}}$

Multiply numerator and denominator by $4$: $\frac{dy}{dx}=\frac{-\frac{1}{\sqrt3}-\sqrt{15}}{2\sqrt3+\frac{2}{\sqrt{15}}}$

Now simplify by writing numerator: $-\frac{1}{\sqrt3}-\sqrt{15}= -\left(\frac{1}{\sqrt3}+\sqrt{15}\right)$

Rationalizing via common structure gives: $\frac{dy}{dx}=-\frac{\sqrt5}{2}$

Match with options

$\frac{dy}{dx}\Big|_{x=1/2}=-\frac{\sqrt5}{2}$

So the correct option is:

B: $-\dfrac{\sqrt5}{2}$