Answer
$$\frac{{dw}}{{dt}} = \frac{{8t - 1}}{{4{t^2} - t + 4}}$$
Work Step by Step
$$\eqalign{
& w = \ln \left( {{x^2} + y} \right),{\text{ }}x = 2t,{\text{ }}y = 4 - t \cr
& \left( {\text{a}} \right){\text{ Calculate }}\frac{{dw}}{{dt}}{\text{ by using the chain rule}} \cr
& \frac{{dw}}{{dt}} = \frac{{\partial w}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial w}}{{\partial y}}\frac{{dy}}{{dt}} \cr
& {\text{Then,}} \cr
& \frac{{\partial w}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {\ln \left( {{x^2} + y} \right)} \right] = \frac{{2x}}{{{x^2} + y}} \cr
& \frac{{\partial w}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {\ln \left( {{x^2} + y} \right)} \right] = \frac{1}{{{x^2} + y}} \cr
& \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {2t} \right] = 2 \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {4 - t} \right] = - 1 \cr
& \underbrace {\frac{{dw}}{{dt}} = \frac{{\partial w}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial w}}{{\partial y}}\frac{{dy}}{{dt}}}_ \downarrow \cr
& \frac{{dw}}{{dt}} = \left( {\frac{{2x}}{{{x^2} + y}}} \right)\left( 2 \right) + \left( {\frac{1}{{{x^2} + y}}} \right)\left( { - 1} \right) \cr
& \frac{{dw}}{{dt}} = \frac{{4x}}{{{x^2} + y}} - \frac{1}{{{x^2} + y}} \cr
& \frac{{dw}}{{dt}} = \frac{{4x - 1}}{{{x^2} + y}} \cr
& {\text{Write in terms of }}t \cr
& \frac{{dw}}{{dt}} = \frac{{4\left( {2t} \right) - 1}}{{{{\left( {2t} \right)}^2} + \left( {4 - t} \right)}} \cr
& \frac{{dw}}{{dt}} = \frac{{8t - 1}}{{4{t^2} - t + 4}} \cr
& \cr
& \left( {\text{b}} \right){\text{By converting }}w{\text{ to a function of }}t{\text{ before differentiating}} \cr
& w = \ln \left( {{x^2} + y} \right),{\text{ }}x = 2t,{\text{ }}y = 4 - t \cr
& w = \ln \left( {4{t^2} + 4 - t} \right) \cr
& \frac{{dw}}{{dt}} = \frac{d}{{dt}}\left[ {\ln \left( {4{t^2} + 4 - t} \right)} \right] \cr
& \frac{{dw}}{{dt}} = \frac{{8t - 1}}{{4{t^2} - t + 4}} \cr} $$
