Hello!! i’m not sure how to do this question, if you could explain your work that’d b great!!

[tex]\bf \sqrt{xy}=y\implies \left( xy \right)^{\frac{1}{2}}=y\implies \stackrel{\textit{chain rule~\hfill }}{\cfrac{1}{2}(xy)^{-\frac{1}{2}}\stackrel{\textit{product rule}}{\left(y+x\cfrac{dy}{dx} \right)}}=\cfrac{dy}{dx} \\\\\\ \cfrac{1}{2\sqrt{xy}}\left(y+x\cfrac{dy}{dx} \right)=\cfrac{dy}{dx}\implies \cfrac{y}{2\sqrt{xy}}+\cfrac{x}{2\sqrt{xy}}\cdot \cfrac{dy}{dx}=\cfrac{dy}{dx}[/tex][tex]\bf \cfrac{x}{2\sqrt{xy}}\cdot \cfrac{dy}{dx}=\cfrac{dy}{dx}-\cfrac{y}{2\sqrt{xy}} \implies \cfrac{x}{2\sqrt{xy}}\cdot \cfrac{dy}{dx}-\cfrac{dy}{dx}=-\cfrac{y}{2\sqrt{xy}} \\\\\\ \stackrel{\textit{common factor}}{\cfrac{dy}{dx}\left( \cfrac{x}{2\sqrt{xy}}-1 \right)}=-\cfrac{y}{2\sqrt{xy}} \implies \cfrac{dy}{dx}=-\cfrac{y}{\left( \frac{x}{2\sqrt{xy}}-1 \right)2\sqrt{xy}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \cfrac{dy}{dx}=-\cfrac{y}{x-2\sqrt{xy}}~\hfill[/tex]