Do "exotic vectorfields" exist?

By an exotic vectorfield on $\mathbb{R}^n$, I mean a non-zero derivation on the algebra $C^0(\mathbb{R}^n)$.

Do such things exist?

No, they don't. Suppose $d:C^0(\mathbb{R}^n)\to C^0(\mathbb{R}^n)$ is a derivation. First, suppose $f\in C^0(\mathbb{R}^n)$ is nonnegative everywhere. Then the nonnegative square root $\sqrt{f}$ is continuous, and $$d(f)=d(\sqrt{f}^2)=2\sqrt{f}d(\sqrt{f}).$$ It follows that if $f(p)=0$, then $d(f)(p)=0$ (since $\sqrt{f}(p)=0$).

Now let $f$ be any continuous function and suppose $f(p)=0$. Then $d(f)=-d(|f|-f)+d(|f|)$. Since $|f|-f$ and $|f|$ are both nonnegative and vanish at $p$, it follows that $d(f)(p)=0$.

But now for any $f$ and any $p$, we can consider the function $g=f-f(p)$. Then $g$ vanishes at $p$, so $d(g)$ vanishes at $p$. But $d(g)=d(f)$ since they differ by a constant. Thus $d(f)$ vanishes at $p$. Since $f$ and $p$ were arbitrary, this means $d=0$.