6. [6] If $x, y, z$ are real numbers such that $x y=6, x-z=2$, and $x+y+z=9$, compute $\frac{x}{y}-\frac{z}{x}-\frac{z^{2}}{x y}$.
Answer:
2
Solution: Let $k=\frac{x}{y}-\frac{z}{x}-\frac{z^{2}}{x y}=\frac{x^{2}-y z-z^{2}}{x y}$. We have
$$
k+1=\frac{x^{2}+x y-y z-z^{2}}{x y}=\frac{x^{2}-x z+x y-y z+z x-z^{2}}{x y}=\frac{(x+y+z)(x-z)}{x y}=\frac{9 \cdot 2}{6}=3,
$$
so $k=2$.