$\sqrt x+\sqrt y=1$
Differentiating on both sides w.r.t $x$ we get,
$\large\frac{1}{2\sqrt x}+\frac{1}{2\sqrt y}\frac{dy}{dx}=$$0$
$\Rightarrow \large\frac{dy}{dx}=-\large\frac{\sqrt y}{\sqrt x}$
$\large\frac{dy}{dx}$ at $\big(\large\frac{1}{4},\frac{1}{4}\big)$ is written as
$\large\frac{dy}{dx}=-\frac{\sqrt{\large\frac{1}{4}}}{\sqrt{\large\frac{1}{4}}}$
$\qquad=-1$