The question asks to write an algorithm using Dijkstra's algorithm with time complexity of
$\Theta(|E| \log |V|)$ that find the second shortest path between $s∈V$ and $t∈V$.
The farthest I managed to get is:
Run Dijkstra's algorithm and find the path from $s$ to $t$. Mark this path as $P_{(s,t)}$.
Remove one edge $e' \in E$ in $P_{(s,t)}$.
Run Dijkstra's algorithm on $G' = (V, E - \{e'\})$ and find the path from $s$ to $t$. Mark this path as $P'_{(s,t)}$.
Define $\text{secondMin} = |P'_{(s,t)}|$.
Define $\text{secondPath} = P'_{(s,t)}$.
For $i = 1$ to $|P_{(s,t)}| - 1$:
a. Remove one edge $e \neq e' \in E$.
b. Run Dijkstra's algorithm on $G'' = (V, E - \{e\})$ and find the path from $s$ to $t$. Mark this path as $P''_{(s,t)}$.
c. if $|P''_{(s,t)}|<$ secondMin then secondMin = $|P''_{(s,t)}|$ and secondPath = $P''_{(s,t)}.$
Return $\text{secondMin}$ and $\text{secondPath}$.
The time complexity of the algorithm is mainly achieved by the loop in line 6:
$$\sum_{i=1}^{\Theta(|V|)} \Theta(|E| \log |V|) = \Theta(|V||E| \log |V|)$$
I will tell you where I got stuck:
It can be shown that each edge on the shortest path between $s$ to $y$, is the shortest edge between two nodes in the path. I can't see the point when Dijkstra's algorithm gives the second shortest path (if it does at all).
In order to get the second shortest path, I know that one green edge on the left need to be removed, but I don't know which one, and that's why I used the loop in line 6.
