First up, partial fractions to make this vaguely approachable.
[math]y=\frac{k x}{1+x}=k-\frac{k}{1-x}=k\left( 1-\frac{1}{1-x}\right)[/math]
Then some substitution: take [imath]u:=\ln(y)[/imath] and [imath]v:=\ln(x)[/imath].
[math]e^{u} = k\left( 1-\frac{1}{1-e^v}\right)[/math]
Take logs.
[math]u = \ln(k) + \ln(1 - \tfrac{1}{1-e^v} )[/math]
Finally, differentiate.
[math]\frac{du}{dv} = \frac{ \tfrac{d}{dv} \left( 1-\tfrac{1}{1-e^v} \right)}{1-\tfrac{1}{1-e^v}}=- \frac{ \tfrac{e^v}{(1-e^v)^2} }{1-\tfrac{1}{1-e^v}}[/math]
Throw [imath]x[/imath] back in and simplify as much as you feel like...
[math]\frac{du}{dv} = \frac{-x}{(1-x)^2 (1-\tfrac{1}{1-x})}= \frac{-x}{x^2 - x}= \frac{-1}{x-1}[/math]
And voilà!
[math]\frac{d\ln(y)}{d\ln(x)}= \frac{-1}{x-1}[/math]
And no, I have no idea why you would want to do that, or why it would be relevant to anything.