Does there exist a continuous function $f:\mathbb R^2\to\mathbb R$ such that $f(0)=0$ and $f(x+y)\neq f(x)+f(y)$ for all linearly independent $x$ and $y$? (On a sidenote: what's a good name for this? My suggestion would be contra-linear or anti-additive; I thought also of anti-linear, nowhere linear and globally non-linear, but these were misleading).

More generally I'm searching for a continuously differentiable function $f:\mathbb R^2\to\mathbb R$ such that $f(x)+f(x+u+v)\neq f(x+u)+f(x+v)$ for all $x$ and all linearly independent $u$ and $v$. Upon failure to exhibit such an example, I sought out to prove the nonexistance of the above family of functions.

(I was lead to this question when investigating whether one can always find the vertices of a parallelogram (or more specifically, a square) in the graph of a continuously differentiable function $f:\mathbb R^2\to\mathbb R$. The nonexistence of anti-additive functions would imply the existence of a parallelogram in the graph of any continuous function.)

(I've edited the original formulation; I forgot the requirement $f(0)=0$.)