Answer
$$\frac{{dU}}{{dt}} = \frac{{1 + 2t + 3{t^2}}}{{t + {t^2} + {t^3}}}$$
Work Step by Step
$$\eqalign{
& {\text{we have }}U = \ln \left( {x + y + z} \right),{\text{ with }}x = t{\text{, }}y = {t^2}{\text{ and }}z = {t^3} \cr
& {\text{find }}\frac{{dU}}{{dt}}{\text{ using the theorem 12}}{\text{.7 }}\left( {{\text{see page 908}}} \right) \cr
& \frac{{dU}}{{dt}} = \frac{{\partial U}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial U}}{{\partial y}}\frac{{dy}}{{dt}} + \frac{{\partial U}}{{\partial z}}\frac{{dz}}{{dt}} \cr
& {\text{find the partial derivatives }}\frac{{\partial U}}{{\partial x}}{\text{,}}\frac{{\partial U}}{{\partial y}}{\text{ and }}\frac{{\partial U}}{{\partial z}} \cr
& \frac{{\partial U}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {\ln \left( {x + y + z} \right)} \right] \cr
& use\,\,\left( {\ln u} \right)' = \frac{1}{u}\left( {u'} \right) \cr
& \frac{{\partial U}}{{\partial x}} = \frac{1}{{x + y + z}}\frac{\partial }{{\partial x}}\left[ {x + y + z} \right] \cr
& \frac{{\partial U}}{{\partial x}} = \frac{1}{{x + y + z}} \cr
& {\text{then similarly}} \cr
& \frac{{\partial U}}{{\partial y}} = \frac{1}{{x + y + z}}{\text{ and }}\frac{{\partial U}}{{\partial z}} = \frac{1}{{x + y + z}} \cr
& \cr
& {\text{find the partial derivatives }}\frac{{dx}}{{dt}}{\text{,}}\frac{{dy}}{{dt}}{\text{ and }}\frac{{dz}}{{dt}} \cr
& \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ t \right] = 1 \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {{t^2}} \right] = 2t \cr
& \frac{{dz}}{{dt}} = \frac{d}{{dt}}\left[ {{t^3}} \right] = 3{t^2} \cr
& \cr
& {\text{substitute the results of the derivatives in the equation of the theorem 12}}{\text{.7}} \cr
& \frac{{dU}}{{dt}} = \frac{{\partial U}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial U}}{{\partial y}}\frac{{dy}}{{dt}} + \frac{{\partial U}}{{\partial z}}\frac{{dz}}{{dt}} \cr
& \frac{{dU}}{{dt}} = \left( {\frac{1}{{x + y + z}}} \right)\left( 1 \right) + \left( {\frac{1}{{x + y + z}}} \right)\left( {2t} \right) + \left( {\frac{1}{{x + y + z}}} \right)\left( {3{t^2}} \right) \cr
& {\text{factoring }}\frac{1}{{x + y + z}} \cr
& \frac{{dU}}{{dt}} = \frac{1}{{x + y + z}}\left( {1 + 2t + 3{t^2}} \right) \cr
& {\text{replace }}x = t{\text{, }}y = {t^2}{\text{ and }}z = {t^3} \cr
& \frac{{dU}}{{dt}} = \frac{1}{{t + {t^2} + {t^3}}}\left( {1 + 2t + 3{t^2}} \right) \cr
& \frac{{dU}}{{dt}} = \frac{{1 + 2t + 3{t^2}}}{{t + {t^2} + {t^3}}} \cr} $$
