Answer
$$\frac{\partial w}{\partial s}=te^{s^2-t^2}+2s^2te^{s^2-t^2},\frac{\partial w}{\partial t}=se^{s^2-t^2}-2st^2e^{s^2-t^2}$$
Work Step by Step
(a) We will use the Chain Rule.
The partial derivative with respect to $s$ is:
$$\frac{\partial w}{\partial s}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial s}+\frac{\partial w}{\partial z}\frac{\partial z}{\partial s}=\frac{\partial }{\partial x}(ze^{xy})\frac{\partial }{\partial s}(s-t)+\frac{\partial }{\partial y}(ze^{xy})\frac{\partial }{\partial s}(s+t)+\frac{\partial }{\partial z}(ze^{xy})\frac{\partial }{\partial s}(st)=ze^{xy}\frac{\partial }{\partial x}(xy)\cdot1+ze^{xy}\frac{\partial }{\partial y}(xy)\cdot1+e^{xy}\cdot t=\
yze^{xy}+xze^{xy}+te^{xy}$$
Expressing this in terms of $s$ and $t$ we have:
$$\frac{\partial w}{\partial s}=yze^{xy}+xze^{xy}+te^{xy}=
(s+t)st\cdot e^{(s-t)(s+t)}+(s-t)st\cdot e^{(s-t)(s+t)}+te^{(s-t)(s+t)}=
s^2t\cdot e^{s^2-t^2}+st^2e^{s^2-t^2}+s^2te^{s^2-t^2}-st^2e^{s^2-t^2}+te^{s^2-t^2}=
2s^2te^{s^2-t^2}+te^{s^2-t^2}$$
The partial derivative with respect to $t$ is:
$$\frac{\partial w}{\partial t}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial t}+\frac{\partial w}{\partial z}\frac{\partial z}{\partial t}=\frac{\partial }{\partial x}(ze^{xy})\frac{\partial }{\partial t}(s-t)+\frac{\partial }{\partial y}(ze^{xy})\frac{\partial }{\partial t}(s+t)+\frac{\partial }{\partial z}(ze^{xy})\frac{\partial }{\partial t}(st)=
ze^{xy}\frac{\partial }{\partial x}(xy)\cdot(-1)+ze^{xy}\frac{\partial }{\partial y}(xy)\cdot1+e^{xy}\cdot s=-yze^{xy}+xze^{xy}+se^{xy}$$
Expressing this in terms of $s$ and $t$ we get:
$$\frac{\partial w}{\partial t}=-yze^{xy}+xze^{xy}+se^{xy}=
-(s+t)ste^{(s-t)(s+t)}+(s-t)ste^{(s-t)(s+t)}+se^{(s-t)(s+t)}=
-s^2te^{s^2-t^2}-st^2e^{s^2-t^2}+s^2te^{s^2-t^2}-st^2e^{s^2-t^2}+se^{s^2-t^2}=
-2st^2e^{s^2-t^2}+se^{s^2-t^2}$$
(b) We will first convert $w$ to a function $s$ and $t$ and then differentiate.
$$w=ze^{xy}=ste^{(s-t)(s+t)}=ste^{s^2-t^2}$$
The partial derivative with respect to $s$ is:
$$\frac{\partial w}{\partial s}=\frac{\partial }{\partial s}(ste^{s^2-t^2)})=te^{s^2-t^2}\frac{\partial }{\partial s}(s)+s\frac{\partial }{\partial s}(te^{s^2-t^2})=te^{s^2-t^2}\cdot1+s\cdot te^{s^2-t^2}\frac{\partial }{\partial s}(s^2-t^2)=te^{s^2-t^2}+ste^{s^2-t^2}\cdot2s=te^{s^2-t^2}+2s^2te^{s^2-t^2}$$
The partial derivative with respect to $t$ is:
$$\frac{\partial w}{\partial t}=\frac{\partial }{\partial t}(ste^{s^2-t^2})=se^{s^2-t^2}\frac{\partial }{\partial t}(t)+t\frac{\partial }{\partial t}(se^{s^2-t^2})=se^{s^2-t^2}\cdot 1+t\cdot se^{s^2-t^2}\frac{\partial }{\partial t}(s^2-t^2)=
se^{s^2-t^2}+ste^{s^2-t^2}\cdot(-2t)=se^{s^2-t^2}-2st^2e^{s^2-t^2}$$
