처음 으로 답글을 적어 봅니다.

결과값은 위와 같이 나옵니다. 참고 하시기 바랍니다.

with t1 (id_user, nm_user, job_user) as
(
select 'tom0' , '톰0', 'doctor' from dual
union all select 'tom1' , '톰1', 'teacher' from dual
union all select 'tom2' , '톰2', 'student' from dual
),
t2 (seq, id_user, weight_user, height_user, dt_reg) as
(
select 1, 'tom0', 70, 170, '20140901' from dual
union all select 1, 'tom1', 80, 180, '20140901' from dual
union all select 1, 'tom2', 90, 190, '20140901' from dual
union all select 2, 'tom0', 71, 171, '20140902' from dual
union all select 2, 'tom1', 81, 181, '20140902' from dual
union all select 2, 'tom2', 91, 191, '20140902' from dual
union all select 3, 'tom0', 72, 172, '20140903' from dual
union all select 3, 'tom1', 82, 182, '20140903' from dual
union all select 3, 'tom2', 92, 192, '20140903' from dual
)
select id_user
, nm_user
, job_user
, seq
, weight_user
, height_user
, dt_reg
from
(
select
t1.id_user
, t1.nm_user
, t1.job_user
, t2.seq
, t2.weight_user
, t2.height_user
, t2.dt_reg
, row_number() over(partition by t2.id_user order by t2.dt_reg desc) RN
from t2, t1
where t1.id_user= t2.id_user
)
where RN = 1
;

--일심님 방법 외 고민을 해봤는데 생각보다 어렵네요. = ㅅ =)ㆀ 
--좋은 방법은 아닌 것 같지만 일심님 데이타셋 이용해서 작성해봅니다.

--#1
SELECT
  *
FROM
  T1, T2 
WHERE
  T1.id_user = T2.id_user 
  AND EXISTS(SELECT 1 FROM T2 T3 GROUP BY id_user HAVING MAX(seq) = T2.seq)
;

--#2
SELECT
  DISTINCT 
  MAX(T1.id_user) OVER(PARTITION BY T1.id_user),
  MAX(T1.nm_user) OVER(PARTITION BY T1.id_user),
  MAX(T1.job_user) OVER(PARTITION BY T1.id_user),
  MAX(T2.seq) OVER(PARTITION BY T1.id_user),
  LAST_VALUE(weight_user) OVER(PARTITION BY T1.id_user ORDER BY T1.id_user),
  LAST_VALUE(height_user) OVER(PARTITION BY T1.id_user ORDER BY T1.id_user)
FROM
  T1, T2
WHERE
  T1.id_user = T2.id_user;

감사합니다

많은 도움이 되었습니다 ^^