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