Answer
$$ y'
=-6 \sinh (9-3t) \cosh (9-3t).$$
Work Step by Step
Since $ y=\cosh^2 (9-3t)$, then the derivative $ y'$, using the chain rule, is given by
$$ y'= 2 \cosh (9-3t) ( \cosh (9-3t))'=2(-3) \sinh (9-3t) \cosh (9-3t) \\
=-6 \sinh (9-3t) \cosh (9-3t).$$
