Update sys.job$ set this_date=:1 where job=:2 as one of the most time-consuming queries - oracle
If that is your most time consuming query, then your database does almost nothing.

That is 0.211599 and 0.002923 seconds per execution with a total runtime of 39.9 and 1.7 seconds.

order by elapsed_time / executions)         )  where rownum &lt;= 10;

This returns the shortest running queries, not the longest-running ones. Use order by ... desc.


No. This is an internal statement that updates the dictionary with the start time of a job on starting it. There is nothing to investigate about it.
You should find this information in  V$SQL_SHARED_CURSOR. Look for the column with a value of Y, and the name of that column specifies the reason for having another child cursor.