I thought this might be quite easy to show, and then realized that the tools I know from real analysis aren't going to help here.
Suppose we have a rational function:
$$ f(X)=\frac{P(X)}{Q(X)} $$
where $Q$ has no linear factors over $\mathbb Q$ (so $Q(x)=0$ if $x\in\mathbb Q$). Now suppose that $f$ is an injection when considered as a function $\mathbb Q\to\mathbb Q$. Must we have $Q=1$; i.e., $f$ is a polynomial?
The only way I could think of to study injectivity algebraically was the following: since $f$ is injective, we have that if $f(x)=f(y)$ then $x=y$. Now treat $x$ and $y$ as invariants; that is to say, take the expression:
$$ \frac{P(X)}{Q(X)}=\frac{P(Y)}{Q(Y)}\\ P(X)Q(Y)=P(Y)Q(X)\\ P(X)Q(Y)-P(Y)Q(X)=0 $$
Now the only solution to $P(x)Q(y)-P(y)Q(x)$ over the rationals is $x=y$ (since $f$ is injective). This means that the expression $P(X)Q(Y)-P(Y)Q(X)$ is the product of $(X-Y)$ and an irreducible; i.e.:
$$ P(X)Q(Y)-P(Y)Q(X)=(X-Y)R(X,Y) $$
where $R\in\mathbb Q[X,Y]$ is irreducible.
I don't really know where to go from here. Is this the right sort of argument?
Isn't $$f(x)=\frac{1}{x^3+3}$$
a counter example?
A "non-trivial" example would be
$$f(x)=\frac{ax^3}{x^3+3}$$
It is easy to see that this is injective when $a \neq 0$ since it can be written as
$$f(x)=a-\frac{3a}{x^3+3}$$
Moreover, as $f(x)$ is decreasing as a real valued function, if $P(x)$ is any polinomial with $P'<0$ then
$$g(x)=P(x)+\frac{1}{x^3+3}$$ satisfies the requirements.