This can be solved with the equation

where A is the new/resulting balance
insert your interest r=0.0146
the starting amount P=9000
the number of times it gets compounded per year n=365 (every day)
and the number of years t=3
insert everything and calculate the balance in 3 years:

the amount earned over 3 years is then the new balance- the current balance:
9402.95-9000=402.95$ earned over 3 years