Answer
$$\,\,\,\,\frac{{dz}}{{dt}} = - \frac{{10}}{3}{t^{7/3}}{e^{1 - {t^{10/3}}}}$$
Work Step by Step
$$\eqalign{
& z = {e^{1 - xy}};\,\,\,\,\,\,\,\,\,x = {t^{1/3}},\,\,\,\,\,\,\,y = {t^3} \cr
& \cr
& {\text{Calculate the partial derivatives }}\frac{{\partial z}}{{\partial x}}{\text{ and }}\frac{{\partial y}}{{\partial x}} \cr
& \,\,\,\,\frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {{e^{1 - xy}}} \right) \cr
& \,\,\,\,{\text{treat }}y{\text{ as a constant}} \cr
& \,\,\,\,\frac{{\partial z}}{{\partial x}} = {e^{1 - xy}}\left( { - y} \right) \cr
& \,\,\,\,\frac{{\partial z}}{{\partial x}} = - y{e^{1 - xy}} \cr
& and \cr
& \,\,\,\,\frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {{e^{1 - xy}}} \right) \cr
& \,\,\,\,{\text{treat }}x{\text{ as a constant}} \cr
& \,\,\,\,\frac{{\partial z}}{{\partial y}} = {e^{1 - xy}}\left( { - x} \right) \cr
& \,\,\,\,\frac{{\partial z}}{{\partial y}} = - x{e^{1 - xy}} \cr
& \cr
& {\text{Calculate the ordinary derivatives }}\frac{{dx}}{{dt}}{\text{ and }}\frac{{dy}}{{dt}} \cr
& \,\,\,\,\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {{t^{1/3}}} \right) = \frac{1}{3}{t^{ - 2/3}} \cr
& \,\,\,\,\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {{t^3}} \right) = 3{t^2} \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{{dx}}{{dt}} + \frac{{\partial z}}{{\partial y}}\frac{{dy}}{{dt}} \cr
& \,\,\,\,\,{\text{Substitute the derivatives}} \cr
& \,\,\,\,\frac{{dz}}{{dt}} = \left( { - y{e^{1 - xy}}} \right)\left( {\frac{1}{3}{t^{ - 2/3}}} \right) + \left( { - x{e^{1 - xy}}} \right)\left( {3{t^2}} \right) \cr
& \,\,\,\,{\text{Where }}x = {t^{1/3}}{\text{ and }}\,\,\,\,\,y = {t^3} \cr
& \,\,\,\,\frac{{dz}}{{dt}} = \left( { - {t^3}{e^{1 - \left( {{t^{1/3}}} \right)\left( {{t^3}} \right)}}} \right)\left( {\frac{1}{3}{t^{ - 2/3}}} \right) + \left( { - {t^{1/3}}{e^{1 - \left( {{t^{1/3}}} \right)\left( {{t^3}} \right)}}} \right)\left( {3{t^2}} \right) \cr
& \,\,\,\,\,{\text{simplifying}} \cr
& \,\,\,\,\frac{{dz}}{{dt}} = - \frac{1}{3}{t^{7/3}}{e^{1 - {t^{10/3}}}} - 3{t^{7/3}}{e^{1 - {t^{10/3}}}} \cr
& \,\,\,\,\frac{{dz}}{{dt}} = - \frac{{10}}{3}{t^{7/3}}{e^{1 - {t^{10/3}}}} \cr} $$
