[docs] Misleading documentation for __prepare__ (issue 15243)

[wkschwartz at gmail.com]

Reviewers: ,

Message:
This looks good and solves my original concern.



Please review this at http://bugs.python.org/review/15243/

Affected files:
  Doc/reference/datamodel.rst


diff -r 7efe1a5239e3 Doc/reference/datamodel.rst
--- a/Doc/reference/datamodel.rst	Fri Jun 17 11:13:03 2016 +0300
+++ b/Doc/reference/datamodel.rst	Fri Jun 17 16:26:41 2016 +0530
@@ -1687,7 +1687,9 @@
 Once the appropriate metaclass has been identified, then the class namespace
 is prepared. If the metaclass has a ``__prepare__`` attribute, it is called
 as ``namespace = metaclass.__prepare__(name, bases, **kwds)`` (where the
-additional keyword arguments, if any, come from the class definition).
+additional keyword arguments, if any, come from the class definition). The
+``__prepare__`` method should be implimented as :func:`classmethod`. Observe
+:ref:`metaclass-example` for demonstration.
 
 If the metaclass has no ``__prepare__`` attribute, then the class namespace
 is initialised as an empty :func:`dict` instance.
@@ -1744,6 +1746,8 @@
       Describes the implicit ``__class__`` closure reference
 
 
+.. _metaclass-example:
+
 Metaclass example
 ^^^^^^^^^^^^^^^^^