Answer
$$ y'
= \frac{6x ( \sinh^{-1} ( x^2))^2}{\sqrt{1+(x^2)^2}}.$$
Work Step by Step
Since $ y=( \sinh^{-1} ( x^2))^3$, then the derivative, by using the chain rule, is given by
$$ y'= 3 ( \sinh^{-1} ( x^2))^2( \sinh^{-1} ( x^2))'=3 ( \sinh^{-1} ( x^2))^2 \frac{(x^2)'}{\sqrt{1+(x^2)^2}}\\
= \frac{6x ( \sinh^{-1} ( x^2))^2}{\sqrt{1+(x^2)^2}}.$$
