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