You can use regular expressions and the word boundary special character \b (highlight by me):

Matches the empty string, but only at the beginning or end of a word. A word is defined as a sequence of alphanumeric or underscore characters, so the end of a word is indicated by whitespace or a non-alphanumeric, non-underscore character. Note that \b is defined as the boundary between \w and \W, so the precise set of characters deemed to be alphanumeric depends on the values of the UNICODE and LOCALE flags. Inside a character range, \b represents the backspace character, for compatibility with Python’s string literals.

def string_found(string1, string2):
    if re.search(r"\b" + re.escape(string1) + r"\b", string2):
        return True
    return False

Demo

If word boundaries are only whitespaces for you, you could also get away with pre- and appending whitespaces to your strings:

def string_found(string1, string2):
    string1 = " " + string1.strip() + " "
    string2 = " " + string2.strip() + " "
    return string2.find(string1)

If you want to make sure that it counts a full word like is will only have one in this is even if there is an is in this, you can split, filter and count:

>>> s = 'this is a sentences that has is and is and is (4)'
>>> word = 'is'
>>> counter = len([x for x in s.split() if x == word])
>>> counter
4

However, if you just want count all occurrences of a substring, ie is would also match the is in this then:

>>> s = 'is this is'
>>> counter = len(s.split(word))-1
>>> counter
3

in other words, split the string at every occurrence of the word, then minus one to get the count.

Edit - JUST USE COUNT:

It's been a long day so I totally forgot but str has a built-in method for this str.count(substring) that does the same as my second answer but way more readable. Please consider using this method (and look at other people's answers for how to)

What is wrong with:

if word in mystring: 
   print('success')

As per @Amadan's comment,

It depends more on the interpretation of the question. Is the word "car" found in the sentence "I like your scarf"? Some people will say yes — there is "car" inside "scarf" — and for those people, your code works correctly.

For this, your code works perfectly fine.

a = input("Enter a sentence:")
w = input ("Enter a word to be found in the sentence:")
if w.lower() in a.lower():
   print("It is present")
else:
   print("It is not present")

Output:

Enter a sentence:I like your scarf
Enter a word to be found in the sentence:car
It is present
>>> 

Some people will say no — "car" is not a word in that sentence — and they would judge your code as incorrect.

For this, use:

a = input("Enter a sentence:")
w = input ("Enter a word to be found in the sentence:")
if w.lower() in a.lower().split():
   print("It is present")
else:
   print("It is not present")

output:

Enter a sentence:I like your scarf
Enter a word to be found in the sentence:car
It is not present

Split up the string into words then test for the substring in each of the words.

For word in s.split():
  If q in word:
    Print word

I'm not sure if I 100% understand what you're trying in a full context but the code below gives the intended results for the example you gave, it returns True if any of the words in the words list are matched with any of the words in the string. Then simply print out the string if result of all is True.

string = 'Vet approved home made dog food'
words = ['dog', 'food']
res = all(word in string for word in words)
if res:
    print(string)
else:
    print('Words not found in string.')

a = """
!
interface blah
a
ssid test1
v
ssid test2
v
ssid test3
"""
p = r'(?<=ssid )\S+' # non-whitespace character chunk after ssid
match = re.findall(p, a)

This will give you: ['test1', 'test2', 'test3']