How can I subtract 3 hours from a datetime in a Pandas dataframe column?

You can use timedelta

from datetime import timedelta

df['startdate'] = pd.to_datetime(df['startdate']) - timedelta(hours=3)
df['enddate'] = pd.to_datetime(df['enddate']) - timedelta(hours=3)

I believe you need convert columns to_datetime and subtract 3 hours timedelta:

cols = ['startdate','enddate']
td = pd.Timedelta(3, unit='h')
df[cols] = df[cols].apply(lambda x: pd.to_datetime(x, format='%d/%m/%Y %H:%M:%S') - td

If want aplly solution for each column separately:

td = pd.Timedelta(3, unit='h')
df['startdate'] = pd.to_datetime(df['startdate'], format='%d/%m/%Y %H:%M:%S') - td
df['enddate'] = pd.to_datetime(df['enddate'], format='%d/%m/%Y %H:%M:%S') - td

print (df)
    cpf  day           startdate             enddate
0  1234    1 2018-01-08 09:50:00 2018-01-08 12:30:00
1  1234    1 2018-01-08 11:30:00 2018-01-08 12:40:00
2  1234    1 2018-01-08 11:50:00 2018-01-08 12:50:00
3  1234    2 2018-02-08 17:20:00 2018-03-07 21:50:00
4  1234    3 2018-03-07 22:00:00 2018-03-08 00:50:00
5  1235    1 2018-01-08 08:50:00 2018-01-08 12:20:00
6  5212    1 2018-01-08 11:50:00 2018-01-08 12:20:00

Last if need convert datetimes to custom format:

df['startdate'] = df['startdate'].dt.strftime('%d/%m/%Y %H:%M:%S')
df['enddate'] = df['enddate'].dt.strftime('%d/%m/%Y %H:%M:%S')
print (df)
    cpf  day            startdate              enddate
0  1234    1  08/01/2018 09:50:00  08/01/2018 12:30:00
1  1234    1  08/01/2018 11:30:00  08/01/2018 12:40:00
2  1234    1  08/01/2018 11:50:00  08/01/2018 12:50:00
3  1234    2  08/02/2018 17:20:00  07/03/2018 21:50:00
4  1234    3  07/03/2018 22:00:00  08/03/2018 00:50:00
5  1235    1  08/01/2018 08:50:00  08/01/2018 12:20:00
6  5212    1  08/01/2018 11:50:00  08/01/2018 12:20:00