Naive approach : Do BFS from every non visited node and keep track of farthest node which is reachable (let's say it's pathlength as COUNT) & mark every intermediate node visited. take maximum of all COUNT and print path from node(which is giving max COUNT) to Farthest node. I think it should work :)

Find a topological sort of the given graph and do dp on it. dp[u] is the length of the longest path which ends in node u. Initially dp[u] = 0 for every u. To calculate this dp go from left to right in topological order and if the current node is u, for every node v such that there is edge from u to v dp[v] = max(dp[v], dp[u] + 1). Then the length of the longest path in the graph is the maximal element in the dp array. To restore the longest path find any u such that dp[u] = max and go through reverse edges of the graph to any v such that dp[v] = dp[u] — 1, until dp[u] is not zero and add u to your answer array. Reverse the answer and it will be one of the possible longest paths in the given DAG. Complexity is O(N + M).

You can practice the same concept on atcoder — https://atcoder.jp/contests/dp/tasks/dp_g

do simple dfs on nodes and store longest path in a dp