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