Show that $\frac{x}{1-x}\cdot\frac{y}{1-y}\cdot\frac{z}{1-z} \ge 8$.

Write $(1-x)=a, (1-y)=b \text { and} (1-z)=c$

$x=2-(y+z)=b+c$

$y=2-(z+x)=a+c$

$z=2-(x+y)=a+b$

Thus we have the same expression in simpler form:

$\dfrac{b+c}{a} \cdot \dfrac{a+c}{b} \cdot \dfrac{a+b}{c}$

Now we have AM-GM:

$b+c \ge 2 \sqrt{bc}$

$a+c \ge 2 \sqrt{ac}$

$b+a \ge 2 \sqrt{ba}$

$\dfrac{b+c}{a} \cdot \dfrac{a+c}{b} \cdot \dfrac{a+b}{c} \ge \dfrac{2^3 abc}{abc} =8$, Done.

Two proofs of the inequality have been posted. The following is simply a comment. In the attempted solution of the OP, the inequality $$\left(\frac{x}{1-x}\cdot\frac{y}{1-y}\cdot \frac{z}{1-z}\right)^{1/3}\ge \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-3}\tag{1}$$ has been proved.

One can complete things by showing, under the hypothesis $x+y+z=2$, that the right-hand side of (1) is $\ge 2$.

However, that is not true. For instance, take $x=\frac{5}{6}$, $y=\frac{5}{6}$, and $z=\frac{2}{6}$. Then $$ \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-3}=\frac{18}{10}\lt 2.$$

So too much has been given away in producing (1): there is probably no direct path from it to the desired result.