Answer
$$\frac{dw}{dt}=6t^2-3$$
Work Step by Step
(a) We will use the Chain Rule:
$$\frac{dw}{dt}=\frac{\partial w}{\partial x}\frac{dx}{dt}+\frac{\partial w}{\partial y}\frac{dy}{dt}+\frac{\partial w}{\partial z}\frac{dz}{dt}=\frac{\partial}{\partial x}(xy+xz+yz)\frac{d}{dt}(t-1)+\frac{\partial}{\partial y}(xy+xz+yz)\frac{d}{dt}(t^2-1)+\frac{\partial}{\partial z}(xy+xz+yz)\frac{d}{dt}(t)=(y+z)\cdot1+(x+z)\cdot2t+(x+y)\cdot1=x+2y+z+2t(x+z)$$
Expressing this in terms of $t$ we have:
$$\frac{dw}{dt}=t-1+2(t^2-1)+t+2t(t-1+t)=t-1+2t^2-2+t+4t^2-2t=6t^2-3$$
(b) We will first convert $w$ to a function of $t$ and then differentiate:
$$w=xy+xz+yz=(t-1)(t^2-1)+(t-1)t+(t^2-1)t=t^3-t-t^2+1+t^2-t+t^3-t=2t^3-3t+1$$
$$\frac{dw}{dt}=\frac{d}{dt}(2t^3-3t+1)=2\cdot3t^2-3=6t^2-3$$
