Use complex numbers!
Let $z = x+iy$, then $|z| = \sqrt{x^2+y^2}$ and $z^2 = (x^2-y^2)+2ixy$. Thus, the complex function: $f : z \mapsto z^2/|z|$ can be identified with your function by taking the real part as the first coordinate, and the imaginary part as the second coordinate.
You hav $U = \{z \in \mathbb{C} : 1 < |z|^2 < 2 \}.$ Let $z \in U$, and consider
$|f(z)| = \left|\frac{z^2}{|z|}\right| = \frac{|z|^2}{|z|} = |z|$
It follows that $|z| = |f(z)|.$ Thus, $z \in U \iff f(z) \in U.$ Consider the two facts: $z \in U \implies f(z) \in U$ tells us that $f(U) \subseteq U,$ while $f(z) \in U \implies z \in U$ tells us that $U \subseteq f(U).$ It follows that $U = f(U).$
The function is clearly not injective since $f(\pm 1) = 1.$ In my notation $1 = 1 + 0i \in \mathbb{C},$ which corresponds to $(1,0) \in \mathbb{R}^2$ in your notation.