Answer
$$\frac{\partial w}{\partial s}=3s^2t^2-t^4,\frac{\partial w}{\partial t}=2s^3t-4st^3$$
Work Step by Step
(a) We will use the Chain Rule.
The partial derivative with respect to $s$ is:
$$\frac{\partial w}{\partial s}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial s}+\frac{\partial w}{\partial z}\frac{\partial z}{\partial s}=\frac{\partial }{\partial x}(xyz)\frac{\partial }{\partial s}(s+t)+\frac{\partial }{\partial y}(xyz)\frac{\partial }{\partial s}(s-t)+\frac{\partial }{\partial z}(xyz)\frac{\partial }{\partial s}(st^2)=yz\cdot1+xz\cdot1+xy\cdot t^2=yz+xz+xyt^2$$
Expressing this in terms of $s$ and $t$ we have:
$$\frac{\partial w}{\partial s}=yz+xz+xyt^2=(s-t)st^2+(s+t)st^2+(s+t)(s-t)t^2=s^2t^2-st^3+s^2t^2+st^3+(s^2-t^2)t^2=2s^2t^2+s^2t^2-t^4=3s^2t^2-t^4$$
The partial derivative with respect to $t$ is:
$$\frac{\partial w}{\partial t}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial t}+\frac{\partial w}{\partial z}\frac{\partial z}{\partial t}=\frac{\partial }{\partial x}(xyz)\frac{\partial }{\partial t}(s+t)+\frac{\partial }{\partial y}(xyz)\frac{\partial }{\partial t}(s-t)+\frac{\partial }{\partial z}(xyz)\frac{\partial }{\partial t}(st^2)=yz\cdot1+xz\cdot(-1)+xy\cdot2st=yz-xz+2stxy$$
Expressing this in terms of $s$ and $t$ wwe have:
$$\frac{\partial w}{\partial t}=yz-xz+2stxy=(s-t)st^2-(s+t)st^2+2st(s+t)(s-t)=s^2t^2-st^3-s^2t^2-st^3+2st(s^2-t^2)=-2st^3+2s^3t-2st^3=2s^3t-4st^3$$
(b)We will first convert $w$ to a function of $s$ and $t$ and then we will differentiate.
$$w=xyz=(s+t)(s-t)st^2=(s^2-t^2)st^2=s^3t^2-st^4$$
The partial derivative with respect to $s$ is:
$$\frac{\partial w}{\partial s}=\frac{\partial }{\partial s}(s^3t^2-st^4)=3s^2t^2-t^4$$
The partial derivative with respect to $t$ is:
$$\frac{\partial w}{\partial t}=\frac{\partial }{\partial t}(s^3t^2-st^4)=2s^3t-4st^3$$
