The answer as to the surjectivity of $\alpha$ is **no**. As in algebraic number theory, the simplest way to prove that an element is not a norm is by local considerations. Let us consider 
$$
y=\frac{(x^3-3x-1)(x^3+3x^2-1)}{x^2(x+1)^2},
$$
and ask whether $y$ is a norm of $\mathbb{Q}(x)$. If $y$ were a norm, we would have
$$
y = f(x)f(-x-1)f(\tfrac{1}{x})f(\tfrac{-1}{x+1})f(-\tfrac{x+1}{x})f(-\tfrac{x}{x+1}),
$$
for some $f \in \mathbb{Q}(x)$. Now since $g=x^3-3x-1$ is an irreducible factor of the numerator of $y$, it must appear as a numerator in one of the six factors in the above product representation of $y$ as well (when written in lowest terms), and since $\phi$ permutes the roots of $g$, it must appear in either three or all six of the factors. Letting $\xi$ be one of the zeros of $g$, this would imply that the order of the zero of the function $y$ (say considered as a meromorphic function in the variable $x$) at $x=\xi$ is a multiple of three, however from the formula for $y$ it is clear that the order of the zero at $x=\xi$ equals $1$.