Answer
$$\frac{{dw}}{{dt}} = - 2t\cos \left( {{t^2}} \right)$$
Work Step by Step
$$\eqalign{
& w = 5\cos xy - \sin xz;\,\,\,\,\,x = 1/t,\,\,\,y = t,\,\,\,z = {t^3} \cr
& {\text{Calculate the partial derivatives }}\frac{{\partial w}}{{\partial x}}{\text{,}}\,\,\frac{{\partial w}}{{\partial y}}{\text{ and }}\frac{{\partial w}}{{\partial z}} \cr
& \,\,\,\,\frac{{\partial w}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {5\cos xy - \sin xz} \right) \cr
& \,\,\,\,{\text{treat }}y{\text{ and }}z{\text{ as constants}} \cr
& \,\,\,\,\frac{{\partial w}}{{\partial x}} = - 5y\sin xy - z\cos xz \cr
& \cr
& \,\,\,\,\frac{{\partial w}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {5\cos xy - \sin xz} \right) \cr
& \,\,\,\,{\text{treat }}x{\text{ and }}z{\text{ as constants}} \cr
& \,\,\,\,\frac{{\partial w}}{{\partial y}} = - 5x\sin xy - 0 \cr
& \,\,\,\,\frac{{\partial w}}{{\partial y}} = - 5x\sin xy \cr
& \cr
& \,\,\,\,\frac{{\partial w}}{{\partial z}} = \frac{\partial }{{\partial z}}\left( {5\cos xy - \sin xz} \right) \cr
& \,\,\,\,{\text{treat }}x{\text{ and }}y{\text{ as constants}} \cr
& \,\,\,\,\frac{{\partial w}}{{\partial z}} = 0 - x\cos xz \cr
& \,\,\,\,\frac{{\partial w}}{{\partial z}} = - x\cos xz \cr
& \cr
& {\text{Calculate the ordinary derivatives }}\frac{{dx}}{{dt}}{\text{,}}\,\,{\text{ }}\frac{{dy}}{{dt}}{\text{ and }}\frac{{dz}}{{dt}} \cr
& \,\,\,\,\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {1/t} \right) = - \frac{1}{{{t^2}}} \cr
& \,\,\,\,\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( t \right) = 1 \cr
& \,\,\,\,\frac{{dz}}{{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{{dw}}{{dt}} = \frac{{\partial w}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial w}}{{\partial y}}\frac{{dy}}{{dt}} + \frac{{\partial w}}{{\partial y}}\frac{{dz}}{{dt}} \cr
& \,\,\,\,\,{\text{Substitute the derivatives}} \cr
& \,\,\,\,\frac{{dw}}{{dt}} = \left( { - 5y\sin xy - z\cos xz} \right)\left( { - \frac{1}{{{t^2}}}} \right) + \left( { - 5x\sin xy} \right)\left( 1 \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( { - x\cos xz} \right)\left( {3{t^2}} \right) \cr
& \,\,\,\,{\text{Where }}\,x = 1/t,\,\,\,y = t,\,\,\,z = {t^3} \cr
& \frac{{dw}}{{dt}} = \left( { - 5t\sin \left( 1 \right) - {t^3}\cos {t^2}} \right)\left( { - \frac{1}{{{t^2}}}} \right) + \left( { - \frac{5}{t}\sin \left( 1 \right)} \right)\left( 1 \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( { - \frac{1}{t}\cos {t^2}} \right)\left( {3{t^2}} \right) \cr
& \,\,\,\,\,{\text{simplifying}} \cr
& \,\,\,\,\,\,\frac{{dw}}{{dt}} = \frac{5}{t}\sin \left( 1 \right) + t\cos {t^2} - \frac{5}{t}\sin \left( 1 \right) - 3t\cos \left( {{t^2}} \right) \cr
& \,\,\,\,\,\,\frac{{dw}}{{dt}} = - 2t\cos \left( {{t^2}} \right) \cr} $$
