If you just gotta do this with a regex, try something like (needs Perl version 5.10+):

c:\@Work\Perl\monks>perl -wMstrict -le
"use 5.010;
 ;;
 my $s =
   'ab + (fun1( fun2(  3  +fun3 ( 4+(5+6)    ))) ) + x + f  ( 1 + 2 )'
+;
 print qq{'$s'};
 ;;
 $s =~ s{ ( \( (?: [^()]++ | (?1) )* \) ) }
        { (my $ns = $1) =~ tr/ \t\n\f//d;  $ns; }xmsge;
 print qq{'$s'};
"
'ab + (fun1( fun2(  3  +fun3 ( 4+(5+6)    ))) ) + x + f  ( 1 + 2 )'
'ab + (fun1(fun2(3+fun3(4+(5+6))))) + x + f  (1+2)'
[download]

This uses Perl version 5.10 Extended Patterns  (?1) (essential; see (?PARNO)) and the  ++ "possessive quantifier" (inessential; could use a  (?>pattern) "atomic" group). If you have Perl version 5.14+, the  tr/// expression in the replacement block can simply be
    { $1 =~ tr/ \t\n\f//dr; }
(note the  /r switch) with no need for substitute-on-copy.

I doubt this regex is the most optimal. I'd like to come up with a regex that just matches  \s+ within a nested paren group and replaces with the empty string, but haven't figured this out yet.

Update: This is a little neater, more elegant, but still not quite what I was hoping for:

c:\@Work\Perl\monks\iaw4>perl -wMstrict -le
"use 5.010;
 ;;
 my $s =
   'ab + (fun1( fun2(  3  +fun3 ( 4+(5+6)    ))) ) + x + f  ( 1 + 2 )'
+;
 print qq{'$s'};
 ;;
 $s =~ s{ [(] (?: [^()]++ | (?0) )* [)] }
        { (my $ns = ${^MATCH}) =~ s/\s+//g;  $ns; }xmsgpe;
 print qq{'$s'};
"
'ab + (fun1( fun2(  3  +fun3 ( 4+(5+6)    ))) ) + x + f  ( 1 + 2 )'
'ab + (fun1(fun2(3+fun3(4+(5+6))))) + x + f  (1+2)'
[download]

Note that the  /p modifier is needed here because  ${^MATCH} is used. And again, the replacement expression becomes a bit simpler with the use of the  /r modifier of Perl 5.14+ on the  s/// operation.