One command that detects none letter in an argument

Quite naturally, check with regular expressions:

\documentclass{article}
\usepackage{xparse}

\ExplSyntaxOn

\NewDocumentCommand{\doubleit}{m}
 {
  \projetmbc_doubleit:n { #1 }
 }

\prg_generate_conditional_variant:Nnn \regex_match:nn { nV } { T,F,TF }
\cs_new_protected:Nn \projetmbc_doubleit:n
 {
  \tl_set:Nn \l_tmpa_tl { #1 }
  % get rid of spaces
  \regex_replace_all:nnN { \s } { } \l_tmpa_tl
  \regex_match:nVTF { .. } \l_tmpa_tl
   {% more than one token
    \regex_match:nVTF { \A -[0-9]* \Z } \l_tmpa_tl
     {% we have a negative number
      \int_eval:n { 2 * (\l_tmpa_tl) }
     }
     {
      2(\l_tmpa_tl)
     }
   }
   {% one token
    \regex_match:nVTF { [0-9] } \l_tmpa_tl
     {% it's a single digit number
      \int_eval:n { 2 * \l_tmpa_tl }
     }
     {%
      2\l_tmpa_tl
     }
   }
 }

\ExplSyntaxOff

\begin{document}

$\doubleit{3}$

$\doubleit{-3}$

$\doubleit{x}$

$\doubleit{-x}$

$\doubleit{x-3}$

\end{document}

Add \, if you want in the middle case, but there's no reason to.

Here's a LuaLaTeX-based solution. If the argument of \doubleit -- say, x -- is a single digit, it is multiplied by 2 directly; if it's a single uppercase or lowercase letter, the output is 2x; in all other cases, the output is 2(x). (If x consists of a single character that's neither a letter nor a digit, nothing is returned since we're presumably dealing with a typo...)

\documentclass{article}
\usepackage{luacode} % for "\luaexec" macro
\luaexec{ % Set up a Lua function:
function doubleit ( s ) 
  if string.len ( s ) == 1 then
    if s:find ( "\%d" ) then     % it's a number!
      tex.sprint ( 2*tonumber(s) )
    elseif s:find ( "\%a" ) then % it's a letter! 
      tex.sprint ( "2" .. s ) 
    end % do nothing if neither number nor letter
  else 
    tex.sprint ( "2(" .. s .. ")" ) 
  end
end
}
% Create a LaTeX "wrapper" macro:
\newcommand\doubleit[1]{\directlua{doubleit("#1")}}

\begin{document}
$\doubleit{x}   \quad 
 \doubleit{3}   \quad 
 \doubleit{-x}  \quad 
 \doubleit{x-3} \quad 
 \doubleit{.}$
\end{document}

The case where the single token is a digit is not really specified in the question. I like egreg's approach of actually doubling, but taking the question literally we could go with

\documentclass{article}
\usepackage{xparse}

\ExplSyntaxOn

\NewDocumentCommand{\doubleit}{m}
 {
  \projetmbc_doubleit:n { #1 }
 }
\cs_new_protected:Nn \projetmbc_doubleit:n
 {
   2
   \bool_lazy_and:nnTF
     { \tl_if_single_token_p:n {#1} }
     { \token_if_letter_p:N #1 }
     {#1}
     { (#1) }
 }

\ExplSyntaxOff

\begin{document}

$\doubleit{3}$

$\doubleit{-3}$

$\doubleit{x}$

$\doubleit{-x}$

$\doubleit{x-3}$

\end{document}