If $\sqrt{x}+\sqrt{y}=1$, then $\frac{d y}{d x}$ at $\left(\frac{1}{4}, \frac{1}{4}\right)$ is :

-1

$\sqrt{x}+\sqrt{y}=1 \Rightarrow \frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \cdot \frac{d y}{d x}=0$

$\Rightarrow \frac{d y}{d x}=-\left.\sqrt{\frac{x}{y}} \Rightarrow \frac{d y}{d x}\right|_{\left(\frac{1}{4}, \frac{1}{4}\right)}=-1$

Hence (3) is correct answer.