Several problems with your approach:
The environment variable JAVA_HOME has nothing to do with the mechanism that your operating system uses for finding executable files. For this the PATH variable is considered. The JAVA_HOME variable is sometimes used by other applications looking for Java.
If using the variable JAVA_HOME, it should be set to the parent directory of the bin directory.
Windows uses back slashes for separating directories, not forward slashes. (EDIT: At least, Windows 7 allows forward slashes, but I would suggest to still use back slashes.) Additionally, there is only one separator between "C:" and the the root directory.
The directory that usually contains installed programs is "Program Files", not "ProgramFiles"
Putting it all together you should do the following:
Create the environment variable JAVA_HOME with your Java installation directory as value, e.g. "C:\Program Files\Java\jdk-1.6". This directory must contain the bin directory, which in turn contains the executable files (such as java.exe).
Extend the environment variable PATH by adding ";%JAVA_HOME%\bin" at the end of it.
Both edits should be made in system variables. Now you simply can open a command prompt and check this setup with a "java -version". This should work.
EDIT
If only user variables can be edited in the environment settings, the following will work:
- Add a variable JAVA_HOME to the user variables, as mentioned above.
- Add a (new) variable PATH to the user variables with the content "%PATH%;%JAVA_HOME%\bin".
If you now open a command prompt and check the path variable by simply typing "path", you will see that the variables in the path are properly expanded. Now check with "java -version".