An algebra problem by Jason Chrysoprase

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

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

Let $2^x = 3^y = 6^{-z} = k$ , so $k^{\frac{1}{x}} = 2, k^{\frac{1}{y}} = 3, k^{\frac{1}{-z}} = 6$

Note that $2 \times 3 = 6$ , so

$\begin{aligned} k^{\large \frac{1}{x}} \times k^{\large \frac{1}{y}} &= k^{\large \frac{1}{-z}} \\ k^{\large \frac{1}{x} + \frac{1}{y}} &= k^{\large \frac{1}{-z}} \\ \dfrac{1}{x} + \dfrac{1}{y} &= \dfrac{1}{-z} \end{aligned}$

Thus,

$\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{-z} + \dfrac{1}{z} = 0$