You have two ways to do it
1.
dim FileNameOnly as string = io.path.getfilename(file_name)
Further reading :
MSDN: Path.GetFileName method[
^]
2.
dim file_info as new IO.FileInfo(file_name)
messagebox.show(file_info.Name)
Further reading :
MSDN: FileInfo Class[
^]
personally I prefer option 1 as you dont have to create any additional objects for the job requirements.