Answer
$$\frac{dw}{dt}=4t^3$$
Work Step by Step
(a) We will use Chain Rule:
$$\frac{dw}{dt}=\frac{\partial w}{\partial x}\frac{dx}{dt}+\frac{\partial w}{\partial y}\frac{dy}{dt}+\frac{\partial w}{\partial z}\frac{dz}{dt}=\frac{\partial}{\partial x}(xy\cos z)\frac{d}{dt}(t)+\frac{\partial}{\partial y}(xy\cos z)\frac{d}{dt}(t^2)+\frac{\partial }{\partial z}(xy\cos z)\frac{d}{dt}(\arccos t)=y\cos z\cdot1+x\cos z\cdot2t-xy\sin z\cdot(-\frac{1}{\sqrt{1-t^2}})=y\cos z+2tx\cos z+\frac{xy\sin z}{\sqrt{1-t^2}}$$
Expressing this in terms of $t$ we have:
$$\frac{dw}{dt}=t^2\cos(\arccos t)+2t\cdot t\cos(\arccos t)+\frac{t\cdot t^2\sin(\arccos t)}{\sqrt{1-t^2}}=t^2\cdot t+2t^2\cdot t+\frac{t^3\sqrt{1-\cos^2(\arccos t)}}{\sqrt{1-t^2}}=t^3+2t^3+\frac{t^3\sqrt{1-t^2}}{\sqrt{1-t^2}}=3t^3+t^3=4t^3$$
(b) We will first convert $w$ to a function of $t$ and then differentiate:
$$w=xy\cos z=t\cdot t^2\cos(\arccos t)=t^3\cdot t=t^4$$
$$\frac{dw}{dt}=\frac{d}{dt}(t^4)=4t^3$$
