It certainly would be nice to know the target RDBMS. But this question is asked so often so let's try and list'em all (at least popular ones) side by side.
For SQL Server:
SELECT e.Ename, e.Eage, e.Eadd, e.Ephone, d.dname
  FROM Employee e LEFT JOIN 
(
  SELECT fkEmpid,
         STUFF((SELECT ',' + dname 
                    FROM Department 
                   WHERE fkEmpid = t.fkEmpid 
                     FOR XML PATH('')) , 1 , 1 , '' ) dname
    FROM Department t
   GROUP BY fkEmpid
) d
    ON e.Empid = d.fkEmpid
Here is SQLFiddle demo
For Mysql, SQLite, HSQLDB 2.X:
SELECT e.Ename, e.Eage, e.Eadd, e.Ephone, d.dname
  FROM Employee e LEFT JOIN 
(
  SELECT fkEmpid,
         GROUP_CONCAT(dname) dname
    FROM Department t
   GROUP BY fkEmpid
) d
    ON e.Empid = d.fkEmpid
Here is SQLFiddle demo (MySql)
Here is SQLFiddle demo (SQLite)
For Oracle 11g:
SELECT e.Ename, e.Eage, e.Eadd, e.Ephone, d.dname
  FROM Employee e LEFT JOIN 
(
  SELECT fkEmpid,
         LISTAGG (dname, ',') WITHIN GROUP (ORDER BY dname) dname
    FROM Department t
   GROUP BY fkEmpid
) d
    ON e.Empid = d.fkEmpid
Here is SQLFiddle demo
For PostgreSQL 9.X:
SELECT e.Ename, e.Eage, e.Eadd, e.Ephone, d.dname
  FROM Employee e LEFT JOIN 
(
  SELECT fkEmpid,
         string_agg(dname, ',') dname
    FROM Department t
   GROUP BY fkEmpid
) d
    ON e.Empid = d.fkEmpid
Here is SQLFiddle demo
Output in all cases:
| ENAME | EAGE | EADD | EPHONE |      DNAME |
---------------------------------------------
|     x |   23 |    b |    677 | test,test1 |
|     y |   24 |    h |    809 |      hello |
|     z |   34 |    u |    799 |     (null) |