MySQL 8.0

By using db<>fiddle, you agree to license everything you submit by Creative Commons CC0.

create table users (
user_id int,
subscription_date date,
service varchar(10));

insert into users values 
(001,'2024-01-01','basic'),
(002,'2024-01-01','plus'),
(002,'2023-01-01','plus'),
(003,'2024-01-01','premium'),
(003,'2024-01-01','premium');

Records: 5  Duplicates: 0  Warnings: 0


select * from users;


user_id
subscription_date
service


1
2024-01-01
basic

2
2024-01-01
plus

2
2023-01-01
plus

3
2024-01-01
premium

3
2024-01-01
premium


select user_id,
       subscription_date,
       service
from (  
       select *,
              count(*) over (partition by user_id,subscription_date,service order by user_id asc ) as cnt 
       from users
     ) sub
where cnt > 1;


user_id
subscription_date
service


3
2024-01-01
premium

3
2024-01-01
premium


select user_id,service,subscription_date
from users
group by user_id,service,subscription_date
having count(*) > 1;


user_id
service
subscription_date


3
premium
2024-01-01

user_id	subscription_date	service
1	2024-01-01	basic
2	2024-01-01	plus
2	2023-01-01	plus
3	2024-01-01	premium
3	2024-01-01	premium