∴ There must be f (0) = 0;

And ∵f(x) is a function with a period of 3, and its domain is r,

∴ There must be f (3k) = 0; K is an integer.

When x∈(0, 1.5), let f(x)=ln(x2-x+ 1)=0.

Then: x2-x+1=1;

The solution is x=0 or x= 1. There are 1 zeros at x∈(0, 1.5).

If f(x) is odd function, the interval x ∈ (-1 .5,0) has1zeros.

According to periodicity, the interval x ∈ (1 .5,3) has1zero points, and the interval x ∈ (1.5,3) has two zero points.

The interval x ∈ (0 0,3) has three zeros.

Then, there are 3×2=6 zeros in the interval (0,6); Then the number of zeros in the interval is 6+ 1=7.