Proof:

Let's start by expressing the given equations in terms of exponents.

Given:
2^x = 3^y = 6^-z

Expressing in terms of exponents:
2^x = 3^y = (2*3)^-z
2^x = 3^y = 2^-z * 3^-z

Equation 1:
2^x = 2^-z * 3^-z

Equation 2:
3^y = 2^-z * 3^-z

To prove: 1/x * 1/y = -1/z

Proof:

We need to prove that 1/x * 1/y = -1/z.

From Equation 1, we can rewrite the equation as:
2^x * 2^z * 3^z = 1

Simplifying the above equation:
2^(x+z) * 3^z = 1

Since any number raised to the power of 0 is equal to 1, we can conclude:
x + z = 0

Similarly, from Equation 2, we can rewrite the equation as:
3^y * 2^z * 3^z = 1

Simplifying the above equation:
3^(y+z) * 2^z = 1

Again, using the property that any number raised to the power of 0 is equal to 1, we can conclude:
y + z = 0

Now, we can solve these two equations simultaneously to find the values of x, y, and z.

From the equation x + z = 0, we can isolate x:
x = -z

From the equation y + z = 0, we can isolate y:
y = -z

Therefore, we can conclude that x = y = -z.

Now, let's prove that 1/x * 1/y = -1/z.

We know that x = y = -z, substituting these values into the equation:
1/x * 1/y = 1/(-z) * 1/(-z)

Simplifying the above equation:
1/x * 1/y = (-1/z) * (-1/z)

Using the property that the product of two negative numbers is positive, we can conclude:
1/x * 1/y = 1/z * 1/z

Simplifying further:
1/x * 1/y = 1/z^2

Therefore, we have proven that 1/x * 1/y = 1/z^2.