Answer
$$\frac{dw}{dt}=24t^3+8$$
Work Step by Step
(a) We will use the Chain Rule:
$$\frac{dw}{dt}=\frac{\partial w}{\partial x}\frac{dx}{dt}+\frac{\partial w}{\partial y}\frac{dy}{dt}+\frac{\partial w}{\partial z}\frac{dz}{dt}=
\frac{\partial}{\partial x}(xy^2+x^2z+yz^2)\frac{d}{dt}(t^2)+\frac{\partial}{\partial y}(xy^2+x^2z+yz^2)\frac{d}{dt}(2t)+\frac{\partial}{\partial z}(xy^2+x^2z+yz^2)\frac{d}{dt}(2)=
(y^2+2xz)\cdot2t+(2xy+z^2)\cdot2+(x^2+2yz)\cdot0=2t(y^2+2xz)+2(2xy+z^2)$$
Expressing this in terms of $t$ we have:
$$\frac{dw}{dt}=2t((2t)^2+2t^2\cdot2)+2(2t^2\cdot2t+2^2)=2t(4t^2+4t^2)+2(4t^3+4)=16t^3+8t^3+8=24t^3+8$$
(b) We will first convert $w$ to a function of $t$ and then differentiate:
$$w=xy^2+x^2z+yz^2=t^2\cdot(2t)^2+(t^2)^2\cdot2+2t\cdot2^2=4t^4+2t^4+8t=6t^4+8t$$
$$\frac{dw}{dt}=\frac{d}{dt}(6t^4+8t)=6\cdot4t^3+8=24t^3+8$$
