There are two ways to do this. One is to check every candidate permutation of letters in the word to see if the candidate is in your dictionary of words. That's an O(N!) operation, depending on the length of the word.
The other way is to check every candidate word in your dictionary to see if it's contained within the word. This can be sped up by aggregating the dictionary; instead of every candidate word, you check all words that are anagrams of each other at once, since if any one of them is contained in your word, all of them are.
So start by building a dictionary whose key is a sorted string of letters and whose value is a list of the words that are anagrams of the key:
>>> from collections import defaultdict
>>> d = defaultdict(list)
>>> with open(r"c:\temp\words.txt", "r") as f:
for line in f.readlines():
if line[0].isupper(): continue
word = line.strip()
key = "".join(sorted(word.lower()))
d[key].append(word)
Now we need a function to see if a word contains a candidate. This function assumes that the word and candidate are both sorted, so that it can go through them both letter by letter and give up quickly when it finds that they don't match.
>>> def contains(sorted_word, sorted_candidate):
wchars = (c for c in sorted_word)
for cc in sorted_candidate:
while(True):
try:
wc = wchars.next()
except StopIteration:
return False
if wc < cc: continue
if wc == cc: break
return False
return True
Now find all the candidate keys in the dictionary that are contained by the word, and aggregate all of their values into a single list:
>>> w = sorted("mythopoetic")
>>> result = []
>>> for k in d.keys():
if contains(w, k): result.extend(d[k])
>>> len(result)
429
>>> sorted(result)[:20]
['c', 'ce', 'cep', 'ceti', 'che', 'chetty', 'chi', 'chime', 'chip', 'chit', 'chitty', 'cho', 'chomp', 'choop', 'chop', 'chott', 'chyme', 'cipo', 'cit', 'cite']
That last step takes about a quarter second on my laptop; there are 195K keys in my dictionary (I'm using the BSD Unix words file).