Answer
$$ y'= \frac{3}{\sqrt{(3x)^2-1}} . $$
Work Step by Step
Recall that $(\cosh^{-1} x)'=\dfrac{1}{\sqrt{x^2-1}}$
Since $ y= \cosh^{-1} (3x)$, then the derivative, by using the chain rule, is given by
$$ y'= \frac{1}{\sqrt{(3x)^2-1}} (3x)'=\frac{3}{\sqrt{(3x)^2-1}} .$$
