Question: Suppose G is Hamiltonian bipartite graph with $ x , y \in V(G) $. Prove that $G-x-y$ has a perfect matching if and only if $x$ and $y$ are on the opposite sides of the graph.
Proof: $\Rightarrow$ Assume that G-x-y has a perfect matching. Since we also know that G is Hamiltonian, |A| = |B| where A and B are the partitions of the G. Therefore, if x and y are not in the opposite partitions, (WLOG, say A) |A| = |B| - $2$. Therefore, there can not be a perfect matching between inequal partitions.
$\Leftarrow$: To be honest, I do not have any brillant ideas for this part. It is very clear actually, because we have a Hamiltonian cycle, which is $ ({x_1}, {y_1}, {x_2}, {y_2},...,{y_n},{x_1})$, and pulling two vertices out will create a hole between the path,therefore I can imagine the matching. However, I could not solve it rigorously.
Thank you!
HINT for the other direction: Let $C$ be a Hamiltonian cycle of $G$ [which is on an even number of vertices]. Then, as you noted already, $G-x-y$; $x \in A; y \in B$ has a perfect matching if [not necessarily only if but we just need if] $C-x-y$ has a perfect matching.
Observe however, that if $x$ and $y$ are on different sides, then [as $G$ is bipartite] every path from $x$ to $y$ in $G$ has an odd number of edges [as each edge in $G$ goes from one side to the other] and thus an even number of vertices. Thus each of the two paths from $x$ to $y$ in $C$ has an even number of vertices. So, can you use this to see that $C-x-y$ consists of two paths each with an even number of vertices, and thus $C-x-y$ has a perfect matching, and [from the above paragraph] that this suffices.