a^x=b^y=c^z.
a=b^(y/x) and c=b^(y/z) so ac=b^(y(1/x+1/z)=b².
Therefore y(1/x+1/z)=2, 1/x+1/z=2/y. This is not the formula in the question!
Put a=16, b=8, c=4 and x=3, y=4, z=6 we have 16³=8⁴=4⁶(=4096). Note that ac=64 and b²=64.
If the formula in the question were correct, 1/x+1/y=2/z would imply that 1/3+1/4=1/3 which is false. But 1/x+1/z=2/y is 1/3+1/6=1/2 which is true.