1. Let $x, y, z$ be positive real numbers, satisfying $x y+z=(x+z)(y+z)$.
Then the maximum value of $x y z$ is $\qquad$
$-1 . \frac{1}{27}$.
From the given condition, we have $z=z(x+y+z)$.
Then $x+y+z=1$.
Thus, $x y z \leqslant\left(\frac{x+y+z}{3}\right)^{3}=\frac{1}{27}$.
When $x=y=z=\frac{1}{3}$, the equality holds.
Therefore, the maximum value of $x y z$ is $\frac{1}{27}$.