jmc

The equation $f(f(x))=x$ turns into $$|3|3x-1|-1|=x.$$This equation implies $3|3x-1|-1=x$ or $3|3x-1|-1=-x.$

Case 1: $3|3x-1|-1=x.$

If $x\ge \frac{1}{3},$ then $|3x-1|=3x-1,$ so $$3(3x-1)-1=x.$$The solution to this equation is $x=\frac{1}{2}.$

If $x<\frac{1}{3},$ then $|3x-1|=1-3x,$ so $$3(1-3x)-1=x.$$The solution to this equation is $x=\frac{1}{5}.$

Case 2: $3|3x-1|-1=-x.$

If $x\ge \frac{1}{3},$ then $|3x-1|=3x-1,$ so $$3(3x-1)-1=-x.$$The solution to this equation is $x=\frac{2}{5}.$

If $x<\frac{1}{3},$ then $|3x-1|=1-3x,$ so $$3(1-3x)-1=-x.$$The solution to this equation is $x=\frac{1}{4}.$

We can check that $\boxed{\frac{1}{5},\frac{1}{4},\frac{2}{5},\frac{1}{2}}$ all satisfy the given equation.