$\frac {\operatorname d\!y}{\operatorname d\!x}$ for $\sqrt{xy}=1$ To find $\displaystyle \dfrac {\operatorname d\!y}{\operatorname d\!x}$ for $\sqrt{xy}= 1$:

$$\dfrac{\sqrt y}{2\sqrt x}+\dfrac{y'\sqrt x}{2\sqrt y}=0\\
\dfrac{y'\sqrt x}{2\sqrt y}=\dfrac{-\sqrt y}{2\sqrt x}\\
y'=\dfrac{-y}x$$

Is this correct? and can you show me how to get the same answer solving using $\ln$?
$\sqrt{xy}=1\implies xy=1 \implies xy'+y=0 \implies y'=-\dfrac yx$.
So, yeah, this is true.
$$\ln \sqrt {xy} = \ln 1$$
$$\frac{\ln{x}+\ln{y}}{2}=0$$
$$\ln y=-\ln x$$
$$\dfrac{\operatorname d}{\operatorname d\!x} \ln y= - \dfrac{\operatorname d}{\operatorname d\!x}\ln x $$
$$\dfrac{\operatorname d \ln y }{\operatorname d\!y} \dfrac{\operatorname dy}{\operatorname d\!x}  =-\frac{1}{x}$$
$$\dfrac{\operatorname dy}{\operatorname d\!x}  =-\frac{y}{x}$$