The 12th power of all numbers (integers) that are not divisible by 5 end in the digits 1 or 6. So when the 12th powers of such numbers are subtracted from one another the result will always end in 0 (from 2 1's or 2 6's) or 5 (from 1-6 or 6-1). Therefore this result is divisible by 5. If one of the numbers is a multiple of 5 its 12th power will end in 5, so if the other number is not divisible by 5 its 12th power will end in 1 or 6 so the difference cannot be divisible by 5. Since 5 is a factor of 1365, it cannot be generally stated that x^12-y^12 is divisible by 1365 since it has just been shown that if x or y (not both) is divisible by 5, the difference is not divisible by 5 and therefore cannot be divisible by 1365. So the statement is not true for all values of x and y. 1365 factorises: 3*5*7*13.
Numbers not divisible by 7 have differences of 6th powers that are divisible by 7, so differences of 12th powers will also be so divisible. The difference between the 12th powers of any two integers neither of which are divisible by 5 or 7 will be divisible by 35.
There are similar rules of divisibility for 3 and 13, but for the statement to be true constraints have to be applied to x and y, and these will exclude divisibility by the factors of 1365. So the statement is generally false.