Dual Barnette's Conjecture Is every Eulerian triangulated (planar) graph Hamiltonian?


On the other hand we have that:


  Barnette's Conjecture (Open): Every cubic bipartite (3-connected) planar graph is Hamiltonian.


Notice that: Eulerian triangulated planar graphs and cubic bipartite planar graphs are mutual duals.

I would be happy to know of a proof or a counterexample (I have not been able to find either). In the absence of both is this a well known conjecture?

Some additional facts (possibly relevant):


Every Eulerian triangulated planar graph is $3$-colorable (see this).
(Whitney's Theorem) Every $4$-connected triangulated planar graph is Hamiltonian.


Definitions:


Hamiltonian Graph: Undirected Graph with a simple cycle through every vertex.
Eulerian Graph: Connected graph with all degrees even.
Triangulated (planar) Graph: Planar graph with every face a triangle.
Cubic Graph: Graph with every vertex having degree exactly $3$.


Edit: (Thanks@domotorp) Triangulated planar graphs are not necessarily Hamiltonian (e.g. see this)
Maybe I am mistaken but I think it is possible to make the example from https://math.stackexchange.com/a/78835/131224 Eulerian. Just take an octahedron and place 3 new vertices on each of its faces, connected to each other and also to two of the vertices of the face in a circular order. So if the face was $abc$, then add $def$ and connect $d$ to all but $a$, $e$ to all but $b$ and $f$ to all but $c$. The same reasoning works but now every degree is even.