Answer
$$\,\,\,\,{\left. {\frac{{dz}}{{dt}}} \right|_{t = 3}} = 1161$$
Work Step by Step
$$\eqalign{
& z = {x^2}y;\,\,\,\,\,x = {t^2},\,\,\,\,y = t + 7 \cr
& \cr
& {\text{Calculate the partial derivatives }}\frac{{\partial z}}{{\partial x}}{\text{ and }}\frac{{\partial z}}{{\partial y}} \cr
& \,\,\,\,\frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {{x^2}y} \right) \cr
& \,\,\,\,{\text{treat }}y{\text{ as a constant}} \cr
& \,\,\,\,\frac{{\partial z}}{{\partial x}} = 2xy \cr
& \cr
& \,\,\,\,\,\frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {{x^2}y} \right) \cr
& \,\,\,\,{\text{treat }}x{\text{ as a constant}} \cr
& \,\,\,\,\frac{{\partial z}}{{\partial y}} = {x^2} \cr
& \cr
& {\text{Calculate the ordinary derivatives }}\frac{{dx}}{{dt}}{\text{ and }}\frac{{dy}}{{dt}} \cr
& \,\,\,\,\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {{t^2}} \right] = 2t \cr
& \,\,\,\,\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {t + 7} \right] = 1 \cr
& \cr
& {\text{Use the Theorem 13}}{\text{.5}}{\text{.1}}\,\,\left( {{\text{chain rule for derivatives}}} \right){\text{ see page 950}} \cr
& \,\,\,\,\frac{{dz}}{{dt}} = \frac{{\partial z}}{{\partial x}}\,\frac{{dr}}{{dt}} + \frac{{\partial z}}{{\partial y}}\frac{{d\theta }}{{dt}} \cr
& \,\,\,\,\,{\text{Substitute the derivatives}} \cr
& \,\,\,\,\frac{{dz}}{{dt}} = \left( {2xy} \right)\,\left( {2t} \right) + \left( {{x^2}} \right)\left( 1 \right) \cr
& \,\,\,\,\frac{{dz}}{{dt}} = 4txy + {x^2} \cr
& \,\,\,\,{\text{Where }}\,x = {t^2},\,\,\,\,y = t + 7 \cr
& \,\,\,\,\frac{{dz}}{{dt}} = 4t\left( {{t^2}} \right)\left( {t + 7} \right) + {\left( {{t^2}} \right)^2} \cr
& \cr
& {\text{Calculate }}{\left. {\frac{{dz}}{{dt}}} \right|_{t = 3}} \cr
& \,\,\,\,{\left. {\frac{{dz}}{{dt}}} \right|_{t = 3}} = 4\left( 3 \right){\left( 3 \right)^2}\left( {3 + 7} \right) + {\left( 3 \right)^4} \cr
& \,\,\,\,\,{\text{simplifying}} \cr
& \,\,\,\,{\left. {\frac{{dz}}{{dt}}} \right|_{t = 3}} = 1080 + 81 \cr
& \,\,\,\,{\left. {\frac{{dz}}{{dt}}} \right|_{t = 3}} = 1161 \cr} $$
