Answer
$$ h'(z)=(1+\frac{1}{z})\sec(z+\ln z)\tan(z+\ln z).$$
Work Step by Step
Recall that $(\ln x)'=\dfrac{1}{x}$
Recall that $(\sec x)'=\sec x\tan x$.
Since $ h(z)=\sec(z+\ln z)$, then the derivative, using the chain rule, is given by
$$ h'(z)=\sec(z+\ln z)\tan(z+\ln z)(z+\ln z)'=(1+\frac{1}{z})\sec(z+\ln z)\tan(z+\ln z).$$
