Answer
$$\frac{{dz}}{{dt}} = 4{t^3} + 3{t^2}$$
Work Step by Step
$$\eqalign{
& {\text{we have }}z = {x^2} + {y^3},{\text{ with }}x = {t^2}{\text{ and }}y = t \cr
& {\text{find }}\frac{{dz}}{{dt}}{\text{ using the theorem 12}}{\text{.7 }}\left( {{\text{see page 908}}} \right) \cr
& \frac{{dz}}{{dt}} = \frac{{\partial z}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial z}}{{\partial y}}\frac{{dy}}{{dt}} \cr
& {\text{find the partial derivatives }}\frac{{\partial z}}{{\partial x}}{\text{ and }}\frac{{\partial z}}{{\partial y}} \cr
& \frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {{x^2} + {y^3}} \right] = 2x \cr
& \frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {{x^2} + {y^3}} \right] = 3{y^2} \cr
& {\text{find the partial derivatives }}\frac{{dx}}{{dt}}{\text{ and }}\frac{{dy}}{{dt}} \cr
& \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {{t^2}} \right] = 2t \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ t \right] = 1 \cr
& \cr
& {\text{substitute the results of the derivatives in the equation of the theorem 12}}{\text{.7}} \cr
& \frac{{dz}}{{dt}} = \frac{{\partial z}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial z}}{{\partial y}}\frac{{dy}}{{dt}} \cr
& \frac{{dz}}{{dt}} = \left( {2x} \right)\left( {2t} \right) + \left( {3{y^2}} \right)\left( 1 \right) \cr
& \frac{{dz}}{{dt}} = 4xt + 3{y^2} \cr
& {\text{replace }}x = {t^2}{\text{ and }}y = t \cr
& \frac{{dz}}{{dt}} = 4\left( {{t^2}} \right)t + 3{\left( t \right)^2} \cr
& \frac{{dz}}{{dt}} = 4{t^3} + 3{t^2} \cr} $$
