Exponents!

$\begin{aligned} 2^x&=3^y=6^{-z} \\ x\log 2&=y\log 3 =-z\log 6 \\ \implies y&=\frac {x\log 2} {\log 3} \\ z&=-\frac {x \log 2}{\log 6} \end{aligned}$

Therefore,

$\begin{aligned} \frac 1x + \frac 1y + \frac 1 z & = \frac 1 x + \frac {\log 3}{x \log 2}-\frac {\log 6}{x\log 2} \\&= \frac {\log 2 + \log 3-\log 6}{x \log 2}\\ &= \frac {\log 6-\log 6}{x \log 2}\\&= \boxed {0}\end{aligned}$