Problem

Your code has some logic problems. Here is the table test for input 23334 :

In short: your code gives an error and you have omitted this information in the question.

  

IndexError: string index out of range

The same is explained in item 28 of the test table.

And even if the error was disregarded, its function would print 9 on the screen, returning a null value.

Solution

First, let's prevent the error from happening. To verify that the x value is not null, just if x: ... . In this way, our function will be:

def count(x):
    y = 0

    if x:
        ... # Implementado posteriormente

    return y

Since x is null, it returns the value of y , which will be 0. Now, within if , we guarantee that x is not null and that there is a 0 position. even has the number 3:

def count(x):
    y = 0

    if x:
        if x[0] == '3':
            y += 1

        ... # Implementado posteriormente

    return y

If the number is 3, we increment the value of y to 1, but regardless of whether or not it is 3, the value of y must be incremented by the amount of 3 digits in the rest of the string. This was your other mistake: you did not accumulate the current value of y with the value that is returned from the next function calls. For this we do:

def count(x):
    y = 0

    if x:
        if x[0] == '3':
            y += 1

        y += count(x[1:])

    return y

And in this way our function is already complete and functional. If we call it passing as a parameter the value '23334' , the answer will be 3.

  

See working at Repl.it .

If you want to count both numbers 3 and 4, you can do:

def count(x):
    y = 0

    if x:
        if x[0] in ('3', '4'):
            y += 1

        y += count(x[1:])

    return y


print(count("23334"))

  

See working at Repl.it .

The test of this solution of my table I leave as an activity for you to do.

Solution without recursion

For a solution without recursion, just use the count method of string object:

>>> print("23334".count("3"))
3

>>> print("23334".count("4"))
1