Thank you Rishi,
Now I finally got it. For those of you who will encounter this
thread looking to solve the same problem, this is how you do it.
1._ Your default page is determined by the "welcome-file" XML entry in your:
<tomcat_inst>conf\web.xml
configuration file. Notice this is ROOT's web.xml! NOT any of the other ones you will find for each web application.
2._ say your "welcome-file" entry looks like this (after installation)
<welcome-file-list>
<welcome-file>index.html</welcome-file>
<welcome-file>index.htm</welcome-file>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
You should only use the "index.jsp" entry because it is a jsp file what yo will be using to forward the request to the servlet. So comment out the whole previous <welcome-file-list> entry and just leave one that should look like:
<welcome-file-list>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
3._ so, <tomcat_inst>\ROOT\index.jsp simpy looks like:
<html>
<head><title>fielder</title></head>
<body>
<jsp:forward page="/servlet/BackStageServlet" />
</body>
</html>
4._ your compile servlet class should be:
<tomcat_inst>\webapps\ROOT\WEB-INF\classes\BackStageServlet.class
5._ then you should do the servlet-mapping to the url that the servlet engine will use to indetify this servlet like this.
<servlet>
<servlet-name>BackStage</servlet-name>
<servlet-class>BackStageServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>BackStage</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
Hope it helped.