Here’s an O(n) solution using a stack instead of repeated search & replace:

closing_to_opening = {')': '(', ']': '[', '}': '{'}
brackets = input()
acc = []
for bracket in brackets:
    if bracket in closing_to_opening:
        if acc and acc[-1] == closing_to_opening[bracket]:
            acc.pop()
        else:
            acc.append(bracket)
    else:
        acc.append(bracket)
print(''.join(acc))

Haven’t thoroughly thought the problem through (so I’m not 100% confident in the correctness of the solution), but the general intuition here is that pairs of brackets can only match up if they only have other matching pairs of brackets between them. You can deal with matching pairs of brackets on the fly simply by removing them, so there’s actually no need for backtracking.

Golfed, just for fun:

a=[]
[a.pop()if a and a[-1]==dict(zip(')]}','([{')).get(b)else a.append(b)for b in input()]
print(''.join(a))