Answer
$$\frac{{dz}}{{dt}} = - \frac{{2t + 2}}{{{{\left( {{t^2} + 2t} \right)}^2}}} - \frac{{3{t^2}}}{{{{\left( {{t^3} - 2} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}z = \frac{1}{x} + \frac{1}{y},\,\,\,\,\,\,\,\,{\text{Where }}x = {t^2} + 2t,{\text{ and }}y = {t^3} - 2 \cr
& \cr
& a.{\text{ Replacing }}x{\text{ and }}y{\text{ to write }}z{\text{ as a function of }}t{\text{ and differentiate}} \cr
& z = \frac{1}{{{t^2} + 2t}} + \frac{1}{{{t^3} - 2}} \cr
& {\text{Differentiate to obtain }}z'\left( t \right) \cr
& z'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{1}{{{t^2} + 2t}}} \right] + \frac{d}{{dt}}\left[ {\frac{1}{{{t^3} - 2}}} \right] \cr
& z'\left( t \right) = \frac{d}{{dt}}\left[ {{{\left( {{t^2} + 2t} \right)}^{ - 1}}} \right] + \frac{d}{{dt}}\left[ {{{\left( {{t^3} - 2} \right)}^{ - 1}}} \right] \cr
& z'\left( t \right) = - \frac{{2t + 2}}{{{{\left( {{t^2} + 2t} \right)}^2}}} - \frac{{3{t^2}}}{{{{\left( {{t^3} - 2} \right)}^2}}} \cr
& \cr
& b.{\text{ Using the chain rule }}\frac{{dz}}{{dt}} = \frac{{\partial z}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial z}}{{\partial y}}\frac{{dy}}{{dt}} \cr
& {\text{Calculating 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[ {\frac{1}{x} + \frac{1}{y}} \right] = - \frac{1}{{{x^2}}} \cr
& \frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {\frac{1}{x} + \frac{1}{y}} \right] = - \frac{1}{{{y^2}}} \cr
& {\text{Calculating the derivatives }}\frac{{dx}}{{dt}}{\text{ and }}\frac{{dy}}{{dt}} \cr
& \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {{t^2} + 2t} \right] = 2t + 2 \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {{t^3} - 2} \right] = 3{t^2} \cr
& \cr
& {\text{Substitute the derivatives into the formula of the chain rule}} \cr
& \frac{{dz}}{{dt}} = \left( { - \frac{1}{{{x^2}}}} \right)\left( {2t + 2} \right) + \left( { - \frac{1}{{{y^2}}}} \right)\left( {3{t^2}} \right) \cr
& {\text{Write in terms of }}t \cr
& \frac{{dz}}{{dt}} = \left( { - \frac{1}{{{{\left( {{t^2} + 2t} \right)}^2}}}} \right)\left( {2t + 2} \right) + \left( { - \frac{1}{{{{\left( {{t^3} - 2} \right)}^2}}}} \right)\left( {3{t^2}} \right) \cr
& {\text{Simplify}} \cr
& \frac{{dz}}{{dt}} = - \frac{{2t + 2}}{{{{\left( {{t^2} + 2t} \right)}^2}}} - \frac{{3{t^2}}}{{{{\left( {{t^3} - 2} \right)}^2}}} \cr} $$
