Let $S$ be the desired sum. Note that $$\frac{x+1}{xy+x+1}=\frac{xz+z}{xyz+xz+z}=\frac{zx+z}{zx+z+1}=\frac{xyz+yz}{xyz+yz+y}=\frac{yz+1}{yz+y+1}$$$$\implies 2\left(\frac{x+1}{xy+x+1}\right)=2-\frac{1}{zx+z+1}-\frac{y}{yz+y+1}$$Summing this cyclically, gives $2S=6-S\implies S=\boxed{2}$.

Let $A$ denote the original sum. $$\frac{x+1}{xy+x+1}=\frac{yz+1}{yz+y+1}$$So $A=\frac{yz+y+2}{yz+y+1}+\frac{z+1}{zx+z+1}$. Now: $$\frac{yz+y+2}{yz+y+1}=\frac{2zx+z+1}{zx+z+1}\Rightarrow A=\frac{2zx+2z+2}{zx+z+1}\Rightarrow A=\boxed2$$