For your particular graph, here's why it can't be Hamiltonian. Consider the "square corners". If there exists an Hamiltonian cycle, this cycle must go through all of them. In particular it must go through every corner ($a,c,e,g,i,k,n,l$), so you know you have to go on all the edges on both squares (i.e. $(a,b), (b,c), (c,h), \dots, (d,a)$ and $(i,j), (j,k), (k,q), \dots, (o,i)$). There must also be at least two edges in the cycle connecting with the vertex $p$ (the one in the middle), for a total of $18$ edges. Since there are $17$ vertices, an Hamiltonian cycle must contain $17$ edges ; we've just shown you need at least $18$ to connect with every vertex, a contradiction.
The key in the argument is that there are a lot of vertices of degree $2$ in your graph ; that gives a lot of restrictions on the possible Hamiltonian cycles.
Hope that helps,