It is known that for metric graphs the concepts of *Gromov's hyperbolicity* and *strictly positive* Cheeger constant are related. Let us first recall the definition of the Cheeger constant. Let $G$ be a graph with vertex set $V(G)$ and edge set $E(G)$. For a collection of vertices $A \subseteq V(G)$, let $\partial A$ denote the collection of all edges going from a vertex in $A$ to a vertex outside of $A$,
$$
 \partial A := \{ (x, y) \in E : x \in A, y \in V(G) \setminus A \}.
$$


Then the Cheeger constant of G, denoted $h(G)$, is defined as
$$
h(G) := \inf \left\{ \frac{|\partial A|}{|A|} : A \subseteq V(G)\quad\text{and}\quad |A|<\infty \right\}. 
$$


However, neither Gromov's hyperbolicity nor strictly positive Cheeger constant implies the other. For instance, the graph $\mathbb{Z}$ is hyperbolic in the sense of Gromov but its Cheeger constant is zero. On the other hand, the Cayley graph of the group $\mathbb{F}\_{2}\times\mathbb{F}\_2$ has positive Cheeger constant but it is not hyperbolic since $\mathbb{Z}\times \mathbb{Z}$ sits as a subgroup.


Does there exist a *planar graph* with positive Cheeger constant but not hyperbolic?



Gromov's Hyperbolicity and Positive Cheeger Constant in Planar Graphs
If I add the assumption that the given graph has bounded degree, and the same holds for the dual graph, then the answer to your question is no. A positive Cheeger constant implies a linear isoperimetric inequality which in turn implies Gromov hyperbolicity. 


However, if you allow the dual graph to have unbounded degree (say), there are examples. Let $T$ be the regular degree $3$ tree. Note that $h(T) > 0$. Properly embed $T$ in the plane. Let $C\_i$ be an ordering of the complementary regions. Now add a single edge $e\_i$, in the interior of $C\_i$, so that $C\_i$ is divided into two regions, one compact and the other not compact. We add the edge so that the compact region is bounded by a cycle of length $i$. 


Let $T' = T \cup (\cup e\_i)$. The cycles created above prevent $T'$ from being Gromov hyperbolic. However $h(T') \geq h(T)$ as adding edges does not decrease the Cheeger constant.


