Looks like I am on a spree today!

Let ${2}^{x}={3}^{y}={6}^{-z}=k$

Therefore, ${k}^{\dfrac{1}{x}}=2$

${k}^{\dfrac{1}{y}}=3$

${k}^{\dfrac{-1}{z}}=6$

It's a very rare, And Unknown fact that $2 \times 3 =6$

$\rightarrow {k}^{\dfrac{1}{x}} \times {k}^{\dfrac{1}{y}}={k}^{\dfrac{1}{-z}}$

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

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

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

Cheers!

There is another way of doing it.

But I think Mehul's solution is easier.

Let $2^x = 3^y = 6^{-z} = k$

Then, $2^x = k , 3^y = k , 6^{-z} = k.$

Applying log ,

$\log_{2}k = x$ , $\log_{3}k = y$ , $\log_{6}k = -z$

Now ,

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

Substituting the values ,

$\dfrac{1}{x} + \dfrac{1}{y} - \dfrac{1}{(-z)} = \dfrac{1}{\log_{2}k} + \dfrac{1}{\log_{3}k} - \dfrac{1}{\log_{6}k} = \log_{k}2 + \log_{k}3 - \log_{k}{6} = \log_{k}6 - \log_{k}6 = \boxed{\boxed{0}}$

Hope it helps.

Cheers!