Because f(2)= 1, f (x)-f (x-3) > 3f (2).

Because f (xy) = f (x)+f (y), 3f(2)=f(2)+f(4)=f(8).

Therefore, f (x) >: f(x-3)+f(8)=f(8(x-3))

And because f(x) increases from zero to positive infinity.

Therefore, x > 8 (x-3) and x-3 >: 0.

Get 3

I'm not sure if the domain you said "Domain 0 is increasing function at positive infinity" is a closed interval. If it contains 0, the result is 3.

This kind of problem is easy to make mistakes. The mistake is to put x>0, X-3 > 0 forgot. This is because f(x) is determined for the increasing function in the field from 0 to positive infinity. This kind of problem is actually very easy to do. Just make good use of the increase and decrease.