I encountered the following problem. Assume that $f\in C_c^{\infty}(\mathbb{R}^n)$ is smooth compactly supported vector field in $\mathbb{R}^n$. Show or disprove: if $f(y)\cdot y=0$ for all $y\in\mathbb{R}^n$, then $\text{div}f=0$. Intuitively this says that the vector field has only irrotational part since it is always perpendicular to radial vectors. However I was not able to prove it and I'm unsure if $\text{div}f=0$ is even true.
Is there a classification for vector fields to have zero divergence? Should I assume more about $f$ to conclude that $\text{div}f=0$?
For $n=1$, the claim is true since the only $f$ that works is obviously the zero function. For $n\geq 2$, take $$f(y)=\left(y_1y_2,-y_1^2,0,0,\ldots,0\right)\ h(y).$$ where $h$ is a compactly supported bump function so $h(y)=1$ for $y\in K$ for some compact set $K$ with non-empty interior, and $h(y)=0$ for $y\notin K'$ for some compact set $K'\supseteq K$.
Then $$y\cdot f(y)=\big(y_1(y_1y_2)+y_2(-y_1^2)\big)\ h(y)=0$$ but, for $y$ in the interior of $K$, $$\operatorname{div}f(y)=\frac{\partial}{\partial y_1}(y_1y_2)+\frac{\partial}{\partial y_2}(-y_1^2)=y_2+0=y_2.$$