Too many x,y,z

If we were to expand the expression, each term of the first parenthesis would be multiplied by the 3 terms of the other parenthesis.Notice that there would be one step in this process where each term would multiply by a visual reciprocal. What I mean by visual reciprocal is that it is noticeable.

$1+1+1+(...)$

$(...)$ = 6 remaining terms

The next thing to notice is that because x+y+z = 0 , the other 2 multiplications of each term would result in other instances of multiplications by reciprocals.

$\dfrac{(x-y)(x)}{(y-z)(z)}$ = $\dfrac{(x-(x+z))x}{(y-(x+y))z}$ = $\dfrac{(-zx)}{(-xz)}$ = 1

Doing this over and over again will leave us with 9 terms and all of them being 1. Resulting in $\boxed{9}$

Currently I am busy , so I am posting a short solution (Not a solution actually). Substitute $(x,y,z)=(1,\omega,\omega^2)$ where $\omega$ is a complex cube root of unity. On simplification we get that the expression equals $\boxed{9}$