Answer
$$\frac{{dy}}{{dx}} = x{\left( {{{\sin }^{ - 1}}x} \right)^2}\left( {\frac{{3x}}{{\sqrt {1 - {x^2}} }} + 2{{\sin }^{ - 1}}x} \right)$$
Work Step by Step
$$\eqalign{
& y = {x^2}{\left( {{{\sin }^{ - 1}}x} \right)^3} \cr
& {\text{differentiate by the product rule}} \cr
& y = {x^2}\left[ {{{\left( {{{\sin }^{ - 1}}x} \right)}^3}} \right]' + {\left( {{{\sin }^{ - 1}}x} \right)^3}\left( {{x^2}} \right)' \cr
& {\text{use chain rule for }}{\left( {{{\sin }^{ - 1}}x} \right)^3} \cr
& \frac{{dy}}{{dx}} = {x^2}\left[ {3{{\left( {{{\sin }^{ - 1}}x} \right)}^2}} \right]\left( {{{\sin }^{ - 1}}x} \right)' + {\left( {{{\sin }^{ - 1}}x} \right)^3}\left( {2x} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = 3{x^2}{\left( {{{\sin }^{ - 1}}x} \right)^2}\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right) + 2x{\left( {{{\sin }^{ - 1}}x} \right)^3} \cr
& {\text{factor out }}x{\left( {{{\sin }^{ - 1}}x} \right)^2} \cr
& \frac{{dy}}{{dx}} = x{\left( {{{\sin }^{ - 1}}x} \right)^2}\left( {3x\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right) + 2\left( {{{\sin }^{ - 1}}x} \right)} \right) \cr
& \frac{{dy}}{{dx}} = x{\left( {{{\sin }^{ - 1}}x} \right)^2}\left( {\frac{{3x}}{{\sqrt {1 - {x^2}} }} + 2{{\sin }^{ - 1}}x} \right) \cr} $$
