Periodic solution of differential equation $y′=f(y)$ 
Let $f\in C^\infty(\mathbb R^2,\mathbb R^2)$ and $\forall x \in \mathbb R^2$ $f(kx)=k^2f(x)$ for $k\in \mathbb R$
Show that any periodic solution of $y′=f(y)$ is constant.

My attempt :
Let $\lambda \in \mathbb{R}$.
Let $g$ a periodic solution defined by $g_{\lambda}(t) = \lambda g(\lambda t)$.
With Cauchy-Lipshitz theorem $g_{\lambda}$ and $g$ do not intersect for $\lambda \neq 1$ and for $\lambda >1$, if $g(0)>0$ we have $g_{\lambda} > g.$
but I do not see how to continue.
 A: Suppose first that $f(x) \neq 0$ when $x \neq 0$ and consider the map $S^1 \to S^1$ given by restricting $f$ to the set $|x| = 1$ and normalizing it. Because $f(x) = f(-x)$, the degree of this map is even. This means there are no periodic solutions $y$ that wind around origin, since those have degree 1.
If $f(x) = 0$ for some $x \neq 0$ then $f(kx) = 0$ for all $k \in {\bf R}$. So in this case there can't be any periodic solution winding around origin either.
Consider now the map $\pi_C : {\bf R}^2 \setminus \{0\} \to S^1$ given by projecting to the unit circle around origin. The above discussion tells us that $\pi_C \circ y : S^1 \to S^1$ is not surjective. But it is continuous, so its image is connected and therefore an interval in $S^1$. Denote this interval by $[a,b]$ and let $x \in y(S^1)$ be one of the points in the pre-image of $a$, that is $\pi_C x = a$. Denote by $X$ the closure of the connected component containing $x$ in $\pi_C^{-1}(a)$. This is an interval $[c,d]$ lying in the half-ray with the slope $a$. Consider now points $s, t \in y(S^1)$ that are close to $c$ and $d$ respectively and both lying on a half-ray with the slope $a + \epsilon$. Because they belong to the same half-ray, $t = ks$ for some $k \in {\bf R}$ and so the vectors $f(t) = k^2 f(s)$ are linearly dependent. But this impossible since the curve $y$ must have distinct tangent vectors at $s$ and $t$ -- one leans towards the half-ray with slope $a$ and the other away from it.
