This isn't an answer to the complexity of the problem, but it at least shows that the complexity has a chance at being nontrivial: it's an example of a cubic graph that cannot be partitioned into paths and claws.

(source: uci.edu)
Within each of its three lobes, any partition into paths and claws can only use six of the seven edges. The remaining six central edges take the form of a claw with each edge subdivided, which cannot be partitioned into paths and claws.
ETA: The graph shown above is more famous as an example of a cubic graph without a perfect matching. But every cubic graph with a perfect matching has a decomposition into paths (not even using any claws). By König's theorem this includes all cubic bipartite graphs and by Petersen's theorem this includes all bridgeless cubic graphs, answering a question of Joseph Malkevitch in the comments.
The proof is very simple: if M is a perfect matching in a cubic graph, the removal of M leaves a 2-regular graph, that is, a disjoint union of cycles. Orient each cycle arbitrarily, and attach each edge uv of M to the cycle edges that follow u and v in the orientations of their cycles.
In the other direction, if there exists a decomposition into paths, then there exists a perfect matching: the middle edges of each path must be a matching since no two middle edges can share a degree-three vertex.
(Disclaimer: this idea may have already been present in Carsten Thomassen's invited talk at GD 2010, which was about this sort of graph decomposition problem.)
(addition to disclaimer (by Anthony Labarre): the "orientation idea" for going from a perfect matching to a partition into paths appears in this paper by Jünger, Reinelt and Pulleyblank, who attribute it to W. H. Cunningham.)