Homogeneity does not suffice for a map between vector spaces to be linear The following problem is taken from Sheldon Axler's book Linear Algebra Done Right, more precisely Exercise 1. from Chapter 3:

Problem: Give an example of a function $f : \mathbb{R}^2 \to \mathbb{R}$ such that $f(av) = af(v)$ for all $a \in \mathbb{R}$ and all $v \in \mathbb{R}^2$ but $f$ is not linear.

I tried to include things like absolute values and square roots in order to handle the homogeneity, but I did not had any success in constructing such an example yet by doing so.

Do you know such an example? Thank you in advance!
consider the map: $$f:\mathbb{R}^2 \to \mathbb{R}\\ (x,y) \mapsto \begin{cases} 
 x \text{ if } y=0 \\
0 \text{ else }\end{cases} $$
then this is homogenous $f(a(x,y))(f(ax,ay))=\begin{cases}ax \text{ if } y=0 \\ a0\text{ else }\end{cases}$, but not linear, since $$f(1,1)+f(1,-1)=0 +0 = 0 \neq f(2,0) =f((1,1)+(1,-1)).$$
consider the map: $$f:\mathbb{R}^2 \to \mathbb{R}\\ (x,y) \mapsto \begin{cases} 
 x \text{ if } y=0 \\
0 \text{ else }\end{cases} $$
then this is homogenous $f(a(x,y))(f(ax,ay))=\begin{cases}ax \text{ if } y=0 \\ a0\text{ else }\end{cases}$, but not linear, since $$f(1,1)+f(1,-1)=0 +0 = 0 \neq f(2,0) =f((1,1)+(1,-1)).$$