Answer
$y=\ln \sqrt x\\z=\dfrac{1}{\sqrt x}\\z=e^{-y}$
Work Step by Step
We are given that $x=t^2; y=\ln t; z=1/t$
Simplify these given terms.
$y=\ln (t)=\ln [\sqrt {t^2}]=\ln \sqrt x$
From $z=\dfrac{1}{t}$
This gives: $z=\dfrac{1}{\sqrt {t^2}}=x^{-1/2}$
Also, $z=\dfrac{1}{t}$ This gives: $z=\dfrac{1}{e^y} \implies z=e^{-y}$
Therefore, $y=\ln \sqrt x\\z=\dfrac{1}{\sqrt x}\\z=e^{-y}$