Not that Tedious

$\begin{aligned} \omega & = \frac{1}{xy+z-1} + \frac{1}{yz+x-1} + \frac{1}{zx+y-1} \\ & = \frac{1}{xy+(2-x-y)-1} + \frac{1}{yz+(2-y-z)-1} + \frac{1}{zx+(2-z-x)-1} \\ & = \frac{1}{xy-x-y+1} + \frac{1}{yz-y-z+1} + \frac{1}{zx-z-x+1} \\ & = \frac{1}{(1-x)(1-y)} + \frac{1}{(1-y)(1-z)} +\frac{1}{(1-z)(1-x)} \\ & = \frac{1-z+1-x+1-y}{(1-x)(1-y)(1-z)} \\ & = \frac{3-(x+y+z)}{1-(x+y+z)+xy+yz+zx-xyz} \\ & = \frac{3-2}{1-2+\frac{1}{2}-4} \\ & = \frac{1}{\left(-\frac{9}{2}\right)} \\ & = -\frac{2}{9} \\ \\ \Rightarrow -18\omega & = \boxed{4} \end{aligned}$

Employing Vieta's Theorems, let x, y and z be the roots of the equation $P^3 - 2P^2 + \dfrac{P}{2} - 4 = 0, \rightarrow 2P^3 - 4P^2 + P - 8 = 0$ Using $x + y + z = 2,re - write, w = \dfrac{1}{(1 - x)(1 - y)} + \dfrac{1}{(1 - y)(1 - z)} + \dfrac{1}{(1 - z)(1 - x)}$ , $\rightarrow w = \dfrac{(1 - x) + (1 - y) + (1 - z)}{(1 - x)(1 - y)(1 - z)}$ . The cubic equation as above can be substituted P = (1 - t), to give $2t^3 - 2t^2 - t + 9 = 0$ , the new cubic equation having roots (1 - x), (1 - y), and (1 - z), sum of them is 1 and their product $\dfrac{-9}{2}$ which are the numerator and denominators of the expression for w as above. Hence $-18w = 4$ , the answer.