Answer
$$c = \frac{2}{3}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^3},{\text{ }}g\left( x \right) = {x^2} + 1{\text{ }}\underbrace {\left[ {0,1} \right]}_{\left[ {a,b} \right]} \cr
& {\text{Differentiate both functions}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^3}} \right] = 3{x^2} \cr
& g'\left( x \right) = \frac{d}{{dx}}\left[ {{x^2} + 1} \right] = 2x \cr
& {\text{Apply the Extended Mean Value Theorem}} \cr
& \frac{{f'\left( c \right)}}{{g'\left( c \right)}} = \frac{{f\left( b \right) - f\left( a \right)}}{{g\left( b \right) - g\left( a \right)}} \cr
& {\text{,then}} \cr
& \frac{{3{{\left( c \right)}^2}}}{{2\left( c \right)}} = \frac{{{{\left( 1 \right)}^3} - {{\left( 0 \right)}^3}}}{{\left[ {{{\left( 1 \right)}^2} + 1} \right] - \left[ {{{\left( 0 \right)}^2} + 1} \right]}} \cr
& {\text{Simplifying}} \cr
& \frac{{3c}}{2} = \frac{1}{1} \cr
& c = \frac{2}{3} \cr} $$
