Logarithmically differentiate an implicit variable power

For the usual real differentiable power with $y>0$, the relation excludes $x=0$. Taking logarithms gives $x\ln y=y+2\ln|x|$. The source writes $2\ln x$, which is the positive-x branch. Differentiate to obtain $\ln y+(x/y)y\prime=y\prime+2/x$. Hence $(x/y-1)y\prime=2/x-\ln y$, so $\dfrac{dy}{dx}=\dfrac{y(2-x\ln y)}{x(x-y)}$ wherever $x\ne y$.