My gut feeling is to say no because this is a bipartite graph with an unequal partition of vertices $(3,1,3)$. However, I am not sure if the exact logic applies because of the line going through above the center as well. Could that suggest that the partitioning is actually different? I still think no but I just want some verification.
Any help?
Your reasoning is wrong: $(3,1,3)$ is not a bipartition at all.
However, the graph still has no Hamiltonian cycle because it is bipartite and has an odd number of vertices. Any Hamiltonian cycle in a bipartite graph must necessarily have an even number of vertices; given a 2-colouring of the graph, all such cycles will alternate colours.