If you want two objects with the same elements but in a different order to compare equal, then the obvious thing to do is compare sorted copies of them - for instance, for the dictionaries represented by your JSON strings a and b:

import json

a = json.loads("""
{
    "errors": [
        {"error": "invalid", "field": "email"},
        {"error": "required", "field": "name"}
    ],
    "success": false
}
""")

b = json.loads("""
{
    "success": false,
    "errors": [
        {"error": "required", "field": "name"},
        {"error": "invalid", "field": "email"}
    ]
}
""")

>>> sorted(a.items()) == sorted(b.items())
False

... but that doesn't work, because in each case, the "errors" item of the top-level dict is a list with the same elements in a different order, and sorted() doesn't try to sort anything except the "top" level of an iterable.

To fix that, we can define an ordered function which will recursively sort any lists it finds (and convert dictionaries to lists of (key, value) pairs so that they're orderable):

def ordered(obj):
    if isinstance(obj, dict):
        return sorted((k, ordered(v)) for k, v in obj.items())
    if isinstance(obj, list):
        return sorted(ordered(x) for x in obj)
    else:
        return obj

If we apply this function to a and b, the results compare equal:

>>> ordered(a) == ordered(b)
True

If you want to print only the difference in subjson (and not the whole structure from the root), you may want to use recursive request

def json_compare(json1, json2):
    #Compare all keys
    for key in json1.keys():
        #if key exist in json2:
        if key in json2.keys():
            #If subjson
            if type(json1[key]) == dict:
                json_compare(json1[key], json2[key])
            else: 
                if json1[key] != json2[key]:
                    print "These entries are different:"
                    print json1[key]
                    print json2[key]
        else:
            print "found new key in json1 %r" % key
    return True