Answer
$$\eqalign{
& \left( a \right){\text{Critical point }}x = \frac{{\sqrt 2 }}{2} \cr
& \left( b \right)f\left( {\frac{{\sqrt 2 }}{2}} \right) = \frac{{{\pi ^2}}}{{16}}{\text{ }}\left( {{\text{and absolute}}} \right){\text{ maximum}} \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \left( {{{\sin }^{ - 1}}x} \right)\left( {{{\cos }^{ - 1}}x} \right){\text{ on the interval }}\left[ {0,1} \right] \cr
& {\text{Differentiate}} \cr
& f\left( x \right) = \left( {{{\sin }^{ - 1}}x} \right)\left( {{{\cos }^{ - 1}}x} \right)' + \left( {{{\cos }^{ - 1}}x} \right)\left( {{{\sin }^{ - 1}}x} \right)' \cr
& f\left( x \right) = {\sin ^{ - 1}}x\left( { - \frac{1}{{\sqrt {1 - {x^2}} }}} \right) + {\cos ^{ - 1}}x\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right) \cr
& f\left( x \right) = \frac{1}{{\sqrt {1 - {x^2}} }}\left( {{{\cos }^{ - 1}}x - {{\sin }^{ - 1}}x} \right) \cr
& {\text{Setting the derivative equal to zero}}{\text{, we have}} \cr
& {\cos ^{ - 1}}x - {\sin ^{ - 1}}x = 0 \cr
& {\cos ^{ - 1}}x = {\sin ^{ - 1}}x \cr
& {\text{Solving this equation for the given interval we obtain}} \cr
& x = \frac{{\sqrt 2 }}{2} \cr
& {\text{Wich lies in the given interval }}\left[ {0,1} \right].{\text{ That points and the endpoints }} \cr
& {\text{are candidates for the location of absolute extrema:}} \cr
& {\text{Points }}x = \left\{ {0,\frac{{\sqrt 2 }}{2},1} \right\} \cr
& {\text{Evaluating these points}} \cr
& f\left( 0 \right) = \left( {{{\sin }^{ - 1}}0} \right)\left( {{{\cos }^{ - 1}}0} \right) = 0 \cr
& f\left( {\frac{{\sqrt 2 }}{2}} \right) = \left( {{{\sin }^{ - 1}}\frac{{\sqrt 2 }}{2}} \right)\left( {{{\cos }^{ - 1}}\frac{{\sqrt 2 }}{2}} \right) = \frac{{{\pi ^2}}}{{16}} \cr
& f\left( 1 \right) = \left( {{{\sin }^{ - 1}}1} \right)\left( {{{\cos }^{ - 1}}1} \right) = 0 \cr
& \cr
& {\text{The largest result evaluating the critical points is }}f\left( {\frac{{\sqrt 2 }}{2}} \right) = \frac{{{\pi ^2}}}{{16}},{\text{ }} \cr
& {\text{which is the local }}\left( {{\text{and absolute}}} \right){\text{ maximum}} \cr} $$
