6. Let $x, y, z > 0$, satisfying $x+y=xy, x+y+z=xyz$. Then the range of $z$ is ( ).
(A) $(0, \sqrt{3}]$
(B) $(1, \sqrt{3}]$
6. D.

Given $y=\frac{x}{x-1}$, we have
$$
\begin{aligned}
z & =\frac{x+y}{x y-1}=\frac{x y}{x y-1}=\left(1-\frac{1}{x y}\right)^{-1} \\
& =\left(1-\frac{1}{x}+\frac{1}{x^{2}}\right)^{-1} .
\end{aligned}
$$

Since $x>1$, i.e., $0<\frac{1}{x}<1$, we have
$$
\frac{3}{4} \leqslant 1-\frac{1}{x}+\frac{1}{x^{2}}<1 \text {. }
$$

Therefore, $1<z \leqslant \frac{4}{3}$.