[tex]\ln x[/tex] is defined for [tex]x>0[/tex], so for [tex]\ln\dfrac{2x}{\sqrt{2x}}[/tex] to be defined, you have to require [tex]\dfrac{2x}{\sqrt{2x}}>0[/tex].
Since [tex]x\neq0[/tex] in this interval, you can simplify the argument slightly:
[tex]\dfrac{2x}{\sqrt{2x}}=\dfrac{2x}{\sqrt2\sqrt x}=\sqrt2\sqrt x=\sqrt{2x}[/tex]
This means [tex]f(x)[/tex] will be defined whenever [tex]\sqrt{2x}>0[/tex], and this happens for all positive [tex]x[/tex].
Computing the derivative is an exercise in applying the chain rule:
[tex]\dfrac{\mathrm d}{\mathrm dx}f(x)=\dfrac{\dfrac{\mathrm d}{\mathrm dx}\sqrt{2x}}{\sqrt{2x}}=\dfrac{\dfrac{\sqrt2}{2\sqrt x}}{\sqrt{2x}}=\dfrac1{2x}[/tex]
