Consider a set $X\epsilon=\{y^2 - \epsilon^2 x^2 \leq 0\} \subset \mathbb{R}^2, 0<\epsilon <<1$ , i.e. a narrow cone passing through the origin. I would like to prove some properties of $f(X\epsilon)$ where $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ is a real polynomial mapping.
Namely, I would like to prove that for small enough $\epsilon$ in some neighbourhood of $f((0, 0))$ there is an analytic curve passing through $f((0, 0))$ and not intersecting with $f(X\epsilon)$ outside of $f((0, 0))$. By curve selection lemma this is equivalent to the germ of $f(X\epsilon)$ being a germ of proper semialgebraic subset of $\mathbb{R}^2$.
Intuitively this seems to be true: the image of $X\epsilon$ for small $\epsilon$ "should be" a small sector bounded by a pair of analytic curves and we can simply choose any curve outside of it.
However, is it actually true? A colleague told me that a counterexample can be constructed using some ideas from ODE theory, however he never managed to point me in the direction of an exact proof.
Additionally, a) $f$ can be assumed to have finite local multiplicity, i.e. $(0, 0)$ is locally the only preimage of $f(0, 0)$; b) all of the critical points of $f$ can be assumed to be located inside $X\epsilon$.