First off check your code. Don't put single quotes around the $myrow[0] and see what happens:
print (date("l, F dS Y", $myrow[0]));
//NOT
print (date("l, F dS Y", '$myrow[0]')); //<<<<SEE THE QUOTES?
SECOND:
It may be that you are not actually storing a date. It appears from your message that you are storing a string.
MySQL and most dbs store dates as numbers, which they then format for presentation. This allows date arithmetic to be done easily.
If you really are storing
xx/xx/xxxx
You're not storing a date, IMO.
Have you defined the date column of your table as DATE or DATETIME or TIMESTAMP?
Do you INSERT or UPDATE with a statement like
SET datecolumn='2003-02-22'
Just like that (quote, year, hyphen, month, hyphen, day, quote).
You can probably convert your string xx/xx/xxxx to an actual DATE that will then format using the date() function of php, by using the strtotime() function.
$mydate=strtotime('01/31/2003');
MIGHT produce right results. You can check to see whether a conversion was done by checking.
if($mydate===-1){echo 'failed';}
I don't know whether you would confuse days and months with this format, however. 02/04/2003: 2nd April or 4th Feb?
Anyway, once you have $mydate as a DATE, you can then use dateformatting:
print (date("l, F dS Y", $mydate));