Actually I found the pattern accidentally when playing with my calculator during some boring or too easy studies at my school. Firstly, I found out that
3x4 = 12
33x4 = 132
333x4 = 1332
3333x4 = 13332
Later I found out it works for 6 and 9 too,
66x4 = 264
666x4 = 2664
6666x4 = 26664

It doesn't work with any other number. It's funny because it seems like the last digit gives 1/3 of it's value to make a new digit on the left.
I made my first quiz and most of my friends can figure it out:
f(3) = 12
f(33) = 132
f(333) = 1332
f(3333) = 13332
What the heck is f(x)?

Soon I noticed that it happens because 3x4 = 12, so
33333x4 = 120000 + 12000 + 1200 + 120 + 12, the number one carries through all the way, converting all number '2' back into 3.

Then somehow I got the idea to combine these numbers.
36x4 = 144
336x4 = 1344
3336x4 = 13344
First, I thought it fails to work, as well as when using other numbers. Soon, though, I realized that each digit just give 1/3 of its value to the previous digit.
The first funny ruleset that I found is very interesting for me. There is a bag, which walks from the right to the left.
It always wants to carry 1/3 of the value of the digit it meets.

For example, 3669693 x 4 = ...?
The bag first comes to the last digit, i.e. number 3, so it takes 1 of it. So the answer's last digit must be 2 (because 1 is taken away by the bag).
Next, when it meets the number 9, the bag wants to have 3. Since it already have 1, it only take 2, thus the number 9 decreases to 7.
When it meets the next number, 6, the bag only want to have 2 while it already have 3, so it gives 1 to the number 6 and it becomes 7.

This works everytime, if the number consists of only the numbers 0, 3, 6 and 9.

I guess, when I reverse it, people may realize the more complex patterns, but not that it is actually as simple as a multiplication of four.

If you know modular arithmetic (seems like you don't?), you'll note that 4 ≡ 2 * 3^-1 mod 10. 0, 3, 6, and 9 are the only numbers modulo 10 that are divisible by 3, so they're the only ones for whom dividing by 3 modulo 10 is the same as dividing by 3 not-modulo-10. Long story short, here are the other numbers besides 4 with this property:

1 * 1^-1 ≡ 1 mod 10 and (1 * 1^-1 mod 10) = 10 - 9*1/1, so e.g. 1*1111 = 01111
1 * 3^-1 ≡ 7 mod 10 and (1 * 3^-1 mod 10) = 10 - 9*1/3, so e.g. 7*3333 = 23331
2 * 3^-1 ≡ 4 mod 10 and (2 * 3^-1 mod 10) = 10 - 9*2/3, so e.g. 4*6666 = 26664
3 * 3^-1 ≡ 1 mod 10 and (3 * 3^-1 mod 10) = 10 - 9*3/3, so e.g. 1*6666 = 06666
5 * 7^-1 ≡ 5 mod 10 and (5 * 7^-1 mod 10) = 10 - 9*5/7, so e.g. 5*7777 = 38885 <- uhh, not sure what's going on here
1 * 9^-1 ≡ 9 mod 10 and (1 * 9^-1 mod 10) = 10 - 9*1/9, so e.g. 9*9999 = 89991
2 * 9^-1 ≡ 8 mod 10 and (2 * 9^-1 mod 10) = 10 - 9*2/9, so e.g. 8*9999 = 79992
3 * 9^-1 ≡ 7 mod 10 and (3 * 9^-1 mod 10) = 10 - 9*3/9, so e.g. 7*9999 = 69993
4 * 9^-1 ≡ 6 mod 10 and (4 * 9^-1 mod 10) = 10 - 9*4/9, so e.g. 6*9999 = 59994
5 * 9^-1 ≡ 5 mod 10 and (5 * 9^-1 mod 10) = 10 - 9*5/9, so e.g. 5*9999 = 49995
6 * 9^-1 ≡ 4 mod 10 and (6 * 9^-1 mod 10) = 10 - 9*6/9, so e.g. 4*9999 = 39996
7 * 9^-1 ≡ 3 mod 10 and (7 * 9^-1 mod 10) = 10 - 9*7/9, so e.g. 3*9999 = 29997
8 * 9^-1 ≡ 2 mod 10 and (8 * 9^-1 mod 10) = 10 - 9*8/9, so e.g. 2*9999 = 19998
9 * 9^-1 ≡ 1 mod 10 and (9 * 9^-1 mod 10) = 10 - 9*9/9, so e.g. 1*9999 = 09999

Hmm, on second thought I don't know what's going on.