Before continuing, let's note that as x, y, z > 0 we have a, b, c >0 and so we may take square roots of any product or quotient thereof, and do so without considering negative solutions.
{(xy = a),(yz = b),(xz=c):}
From the first equation, we have
y = a/x
Substituting this into the second equation:
b = yz = a/xz
=> z = b/ax
Substituting this into the third equation:
c = xz = b/ax^2
=> x^2 = (ac)/b
=> x = sqrt((ac)/b)
Substituting our solution for x into our intermediate result from working on the first equation:
y =a/x = a/sqrt((ac)/b) = sqrt((ab)/c)
Substituting our solution for x into our intermediate result from working on the second equation:
z = b/ax = b/asqrt((ac)/b) = sqrt((bc)/a)
:.{(x = sqrt((ac)/b)), (y = sqrt((ab)/c)), (z = sqrt((bc)/a)):}