Can the Vertices of cubic graph be partitioned into and induced cycle and a forest? Let $G$ be a $2$-connected $3$-regular graph. Can $V(G)$ be partitioned into  $V_1$ and $V_2$ where 
$G[V_1]$(the induced subgraph on $V_1$) is a cycle of $G$ and $G[V_2]$ is a forest (Acyclic subgraph) of $G$?
Edit:
Since the counterexamples presented so far are $2$-connected, what happen if the connectivity of the graph is $3$ instead of two?
 A: I believe this is false.
EDIT
The previous counterexample was wrong, let me try again.
A program found counterexample on $10$ vertices and exhaustive search
confirmed it.
The edges are:
[(0, 3), (0, 5), (0, 7), (1, 4), (1, 6), (1, 9), (2, 6), (2, 7), (2, 8), (3, 5), (3, 7), (4, 8), (4, 9), (5, 8), (6, 9)]

It is $2$-connected.

According to a search the smallest $3$-connected counterexample
is on $12$ vertices with edges
[(0, 6), (0, 7), (0, 8), (1, 6), (1, 7), (1, 9), (2, 6), (2, 10), (2, 11), (3, 7), (3, 10), (3, 11), (4, 8), (4, 9), (4, 10), (5, 8), (5, 9), (5, 11)]

There are other counterexamples (modulo errors).
A: For a 3-connected counterexample, you can take a 'truncated' Petersen graph, truncating meaning that we remplace each vertex of $Pete$ by a 'small' triangle (think of a truncated cube). Then an induced cycle $G[V_1]$ as above must hit each triangle and moreover alternate between triangle edges and edges of the original $Pete$. Thus the latter edges would form a Hamilton cycle of $Pete$, which doesn't exist.
Note that similar truncating constructions for higher connectivity $k>3$ won't work, as  for $k=4$ a non Hamiltonian 4-regular 4-connected graph just doesn't exist (if my memory serves), and for $k>4$ a 'truncated vertex' $C_k$ might be visited more than once.
Edit after Soro's comment: Truncating the Meredith graph $M$ (i.e. replacing each vertex by a small $C_4$ - note that this can be done in 3 different ways for each vertex, but that isn't relevant) will result in a cubic graph with 280 vertices that is still 4-connected. If a cycle as above exists, it must hit each $C_4$ (sharing either one or two adjacent edges with it) and alternatingly with those, hit edges of $M$, which would again form a Hamilton cycle of $M$.  
Just a feeling: for a $5$-connected graph, I would conjecture the statement to hold.
