Prove that following graph $G$ has no Hamilton circuit. Prove that following graph $G$ has no Hamilton circuit exists.

$\underline{Attempt}$
First I assumed that the graph $G$ has a Hamilton circuit.
then,

*

*Since graph is symmetric, deleted the edge $(j,h)$


*Since G has a Hamilton circuit, At $a$ has an edge either $(c,a)$ or $(a,d)$ so I delete the $(c,a)$


*It follows that $c,a,b,d,j,c$ has subcircuit

Thus which contradicts my assumption so there does not exist Hamilton circuit
Can anyone verify my answer? If it is wrong please correct me?

Your proof seems correct to me, though you deleted $(a,d)$ and not $(c,a)$, and you should really also say that you were allowed to choose either one wlog because of mirror symmetry of the graph, and in the previous step that you could delete $(j,h)$ due to rotational symmetry putting the unused $j$-edge there.
Another way to prove this is to say that if you remove the three vertices $c$, $d$, $h$, and their edges, you get four disconnected components ($\{a,b\}$, $\{g,e\}$, $\{f,i\}$, and $\{j\}$). Any Hamiltonian circuit will have to go through one of the removed vertices to move from one component to the next. But there are four components to be visited and only three vertices to alternate them with so they cannot form a circuit.
