Hmm...II

If $d = \frac{f-g}{f+g}$ , then we can see that $1+d = \frac{2f}{f+g}$ and $1-d = \frac{2g}{f+g}$

So $\dfrac{1+d}{1-d} = \dfrac{\frac{2f}{f+g}}{\frac{2g}{f+g}}=\dfrac{f}{g}$

So with what we are given, we can see that: $\begin{array}{c}~\dfrac{1+x}{1-x}\cdot\dfrac{1+y}{1-y}\cdot\dfrac{1+z}{1-z} &= \dfrac{a}{b}\cdot\dfrac{b}{c}\cdot\dfrac{c}{a}\\~\\~\\&=\boxed{1} \end{array}$