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