This is a good question. Here's a proof one way...hopefully it'll help you do it in reverse too.

We have x^2 + y^2 = 11xy.

Therefore x^2 - 11xy + y^2 = 0.

Note that (x - y)^2 = x^2 - 2xy + y^2.

Therefore (x - y)^2 = 9xy.

Take logs of both sides:

log((x - y)^2) = log(9xy).

Therefore 2log(x-y) = log(9) + log(x) + log(y).

(*) Therefore log(x-y) = 0.5(log(9) + log(x) + log(y)).

Now log((x-y)/3) = log(x-y) - log(3).

Therefore log(x-y) = log((x-y)/3) + log 3.

Putting this into (*):

log((x-y)/3) = -log(3) + 0.5(log(9) + log(x) + log(y)).

Therefore log((x-y)/3) = -log(3) + log(3) + 0.5(log(x) + log(y)).

Done!

Hope this helps.