Answer
$\dfrac{dy}{dx}=\dfrac{y-4\sqrt{(xy)^3}}{2x^2\sqrt{xy}-x}.$
Work Step by Step
$\dfrac{d}{dx}(\sqrt{xy})=\dfrac{d}{dx}(x^2y)+\dfrac{d}{dx}(1)\rightarrow$
Using the Chain Rule with $u=xy\rightarrow =\dfrac{dy}{dx}=(y+\dfrac{dy}{dx}(x))\rightarrow$
$\dfrac{d}{dx}(\sqrt{xy})=\dfrac{y+\dfrac{dy}{dx}(x)}{2\sqrt{xy}}\rightarrow$
$\dfrac{y+\dfrac{dy}{dx}(x)}{2\sqrt{xy}}=(2xy+\dfrac{dy}{dx}(x^2))\rightarrow$
$y-4\sqrt{(xy)^3}=\dfrac{dy}{dx}(2x^2\sqrt{xy}-x)\rightarrow$
$\dfrac{dy}{dx}=\dfrac{y-4\sqrt{(xy)^3}}{2x^2\sqrt{xy}-x}.$
