Let me first answer your last question in the negative: there exist homeomorphisms $f:M \rightarrow M$ of smoothable manifolds $M$ such that neither $f$ nor $f^{-1}$ are smooth with respect to any smooth structure on $M$. In fact, we can take $M = S^3$. Bing has constructed homeomorphisms $f:S^3 \rightarrow S^3$ such that $f \circ f = \text{id}$ and such that the fixed set of $f$ is the Alexander horned sphere. See
Bing, R. H.
A homeomorphism between the 3-sphere and the sum of two solid horned spheres.
Ann. of Math. (2) 56, (1952). 354–362.
The map $f$ (and hence the map $f^{-1} = f$) is not smooth with respect to any smooth structure on $S^3$ because of a theorem of P. Smith that says that the fixed set of any periodic diffeomorphism of $S^3$ is a smoothly embedded surface or circle. This is the beginning of a long story that culminated in the proof of the Smith conjecture, which says that any periodic diffeomorphism of $S^3$ is smoothly conjugate to an element of the orthogonal group. This was one of the early triumphs of Thurston's work; for a detailed account of it, see the book
The Smith conjecture.
Papers presented at the symposium held at Columbia University, New York, 1979. Edited by John W. Morgan and Hyman Bass. Pure and Applied Mathematics, 112. Academic Press, Inc., Orlando, FL, 1984. xv+243 pp. ISBN: 0-12-506980-4
This brings me to my second point. It is FALSE that the map $g:\mathbb{R} \rightarrow \mathbb{R}$ defined via $g(x) = x^3$ is not a diffeomorphism with respect to any smooth structure on $\mathbb{R}$. It is true that $\mathbb{R}$ has a unique smooth structure (as does $S^3$ above), but this means something weaker than what you are claiming it means. Namely, if $X$ and $Y$ are any two smoothings of $\mathbb{R}$, then there exists a diffeomorphism $X \rightarrow Y$; however, this diffeomorphism might not be the identity.
Here is a sketch of how you construct the appropriate smooth structure. Constructing this smooth structure is equivalent to constructing a homeomorphism $\psi:\mathbb{R} \rightarrow \mathbb{R}$ such that $\psi \circ g \circ \psi^{-1}$ is a smooth map; the desired smooth structure is then obtained by pulling back the standard smooth structure on $\mathbb{R}$ via $\psi$. Here are some observations about $g$:
It has three fixed points, namely $-1$ and $0$ and $1$.
On the interval $(-\infty,-1)$, it shifts points to the left.
On the interval $(-1,0)$, it shifts points to the right.
On the interval $(0,1)$, it shifts points to the left.
On the interval $(1,\infty)$, it shifts points to the right.
It is easy to construct a diffeomorphism $h: \mathbb{R} \rightarrow \mathbb{R}$ with the above properties (since $0$ will be an attracting fixed point of $h$, the derivative of $h$ at $0$ will be a number $\alpha$ satisfying $0 < \alpha < 1$). But it is then an easy exercise to see that $h$ will be topologically conjugate to $g$.