Answer
$$y'=-\frac{4(csch^{-1} ( 3x))^3}{|x|\sqrt{(3x)^2-1}}.$$
Work Step by Step
Recall that $(\csc^{-1} x)'=\dfrac{-1}{|x|\sqrt{x^2-1}}$
Since $ y=( csch^{-1} ( 3x))^4$, then the derivative, by using the chain rule, is given by $$ y'= 4 (csch^{-1} ( 3x))^3( csch^{-1} (3 x))'=4 (csch^{-1} ( 3x))^3\frac{-3}{|3x|\sqrt{(3x)^2-1}}\\ =-\frac{12(csch^{-1} ( 3x))^3}{|3x|\sqrt{(3x)^2-1}}=-\frac{4(csch^{-1} ( 3x))^3}{|x|\sqrt{(3x)^2-1}}.$$
