The marking will be at :
1/5,1/3,2/5,3/5,2/3, and 4/5.
The lengths between any 2 successie markings would be
1/5,2/15,1/15,1/5,1/15, and 2/15
Out of these first 3 are unique.So upto 2/5 the lenghts would be unique. remaining is 3/5. Is that the answer?

Let us assume that the length of the branch is 15 units (Why, an assumption ? Because the question asks about 'Fraction of Original Length' which is a relative measure, so an assumption makes solving easy).

If such branch is broken as per the conditions in the question, it will result in

2 pieces of 1 Unit length,
2 pieces of 2 Units length, and
3 pieces of 3 Units length.

If one piece of every distinct length is removed from the branch, then it will result in

1 pieces of 1 Unit length,
1 pieces of 2 Units length, and
2 pieces of 3 Units length.

Length of the remains = (1*1) + (1*2) + (2*3) = 6+2+1 = 9
Original length of the branch = 15
Fraction of the original branch which remains = 9/15 = 3/5.

Please refer to the attachment for better understanding!