How to prove NP-hardness of a longest-path problem?
Assuming I understand your definition, your problem is related to the optimisation version of LONGEST PATH:


  LONGEST PATH
  Input: A graph $G=(V,E)$, an integer $k$.
  Question: Is there a path with at least $k$ vertices in $G$


This problem is NP-complete, there's a fairly obvious reduction from HAMILTONIAN PATH - just set $k = n$, and clearly if we are given an (ordered) set of vertices, we can easily check that it is a path over at least $k$ vertices.

Note that we can assume that $G$ is connected, and we will do so from now on.

The optimization version max-LONGEST PATH where we ask for the path of maximum length is then NP-hard (the polynomial-time equivalence of these two problems is a straightforward exercise).

Then the difference between max-LONGEST PATH and your problem is that you have two specified vertices that are the start and the end of the path. Now we need a reduction from some NP-hard problem to your problem, but we already have one that's really close.

Given an instance $G$ of max-LONGEST PATH you should now be able to construct an instance $G'$ of your problem such that if $p$ is the longest path in $G$, then this is almost the longest path in $G'$.

A further hint if you are still beating your head against a wall in about, say, 2 hours:


   Try taking $G$ and adding two new vertices $s$ and $t$ and connecting them so that they can start and end such a path $p$ - remember though that you don't know what $p$ is.