Answer
$$\left( {{\text{ }}{f^{ - 1}}} \right)'\left( 2 \right) = \frac{6}{{13}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \sqrt {{x^3} + x - 1} ,{\text{ at }}y = 3 \cr
& {\text{For }}y = 3 \cr
& 3 = \sqrt {{x^3} + x - 1} \cr
& {\left( 3 \right)^2} = {\left( {\sqrt {{x^3} + x - 1} } \right)^2} \cr
& 9 = {x^3} + x - 1 \cr
& {x^3} + x - 10 = 0 \cr
& {\text{Solving by using a calculator or a CAS}}{\text{, we obtain}} \cr
& x = 2 \cr
& {\text{Calculate }}f'\left( x \right) \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\sqrt {{x^3} + x - 1} } \right] \cr
& f'\left( x \right) = \frac{{3{x^2} + 1}}{{2\sqrt {{x^3} + x - 1} }} \cr
& {\text{Calculate }}f'\left( 2 \right) \cr
& f'\left( 2 \right) = \frac{{3{{\left( 2 \right)}^2} + 1}}{{2\sqrt {{{\left( 2 \right)}^3} + \left( 2 \right) - 1} }} = \frac{{13}}{6} \cr
& {\text{The derivative of }}{f^{ - 1}}\left( x \right){\text{ is}} \cr
& \left( {{\text{ }}{f^{ - 1}}} \right)'\left( x \right) = \frac{1}{m} = \frac{1}{{f'\left( x \right)}} \cr
& {\text{Therefore}}{\text{,}} \cr
& {\text{ }}\left( {{\text{ }}{f^{ - 1}}} \right)'\left( 2 \right) = \frac{1}{{f'\left( 2 \right)}} \cr
& \left( {{\text{ }}{f^{ - 1}}} \right)'\left( 2 \right) = \frac{1}{{13/6}} \cr
& \left( {{\text{ }}{f^{ - 1}}} \right)'\left( 2 \right) = \frac{6}{{13}} \cr} $$
