Answer
$$y'=-\frac{y}{x}$$
Work Step by Step
$$\sqrt{xy}=1$$
$$(xy)^{1/2}=1$$
Using implicit differentiation, we differentiate both sides of the equation with respect to $x$: $$\frac{1}{2}(xy)^{-1/2}\times(xy)'=0$$
$$\frac{1}{2}(xy)^{-1/2}\times(y+xy')=0$$
$$\frac{1}{2}(xy)^{-1/2}\times y+\frac{1}{2}(xy)^{-1/2}\times(xy')=0$$
$$\frac{y}{2\sqrt{xy}}+\frac{x}{2\sqrt{xy}}\times y'=0$$
Now we set all elements with $y'$ to one side and the rest to the other side: $$\frac{x}{2\sqrt{xy}}\times y'=-\frac{y}{2\sqrt{xy}}$$
Finally, we calculate $y'$: $$y'=-\frac{\frac{y}{2\sqrt{xy}}}{\frac{x}{2\sqrt{xy}}}=-\frac{y\times2\sqrt{xy}}{x\times2\sqrt{xy}}$$ $$y'=-\frac{y}{x}$$
