f'(x)= 1/x-2a/x2

g(x)= 1/x-2a/x2-x/6 =-(-6x+ 12a+x3)/(6x 2)

Let h(x)=x3-6x+ 12a.

From h'(x)=3x2-6=0, x=√2.

h(0)= 12a，

H(√2)=-4√2+ 12a is the minimum value of h(x).

Discussion a

When a≤0, h(x) has only one positive zero x >: √2, so g(x) has only 1 zero;

When 0; √2, so g(x) also has two zeros;

When a=√2/3, h(x) has only 1 positive zeros, while g(x) has only 1 positive zeros.

When a & gt√2/3, h(x) has no positive zero, and g(x) has no zero at this time.