It is not possible and here is why:
If the graph is traceable, all vertices must have an even number of connections from it, except for possibly two of them. This is because you must both arrive and leave at every point except for the start and the end. You may visit each point any number of times, but since you must both arrive and leave (2 connections) each time you visit a vertex, the total number of connections for a transversed vertex is
2*(# of visits) = # of connections, an even number
If the start and the end are on different vertices, those two would have an odd number. This is a quick method of finding the start and finish vertices, by locating the two vertices with an odd number of connections. If the graph has more than two vertices with an odd number, then is not traceable. If one vertex is the start and a second is the end, then the others must be transversed and the above formula applies to the number of connections. For the formula to give the correct number of connections for those other odd vertices, we must have visited them a fractional number of times, which is not possible.
In the example, we have four vertices with an odd number of connections. These four points have 3, 3, 7, and 5 connections. If the two 3’s are the start and end, then we must have visited the vertex with 7 connections 3.5 times and the vertex with 5 connections 2.5 times, a logical impossibility.