If we define $F(x,y,z)\equiv f_1(x)+f_2(y)+f_3(z)$, the curl of $\vec{f}(\vec{r})=F(x,y,z)\vec{r}=(F\cdot x,F\cdot y,F\cdot z)$ would be
\begin{align*} \nabla\times\vec{f}&= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \partial_x & \partial_y & \partial_z \\ Fx & Fy & Fz \end{vmatrix}\\ &=(z\partial_y F-y\partial_z F,x\partial_z F-z\partial_x F,y\partial_x F-x\partial_y F)\\ &= (zf_2'-yf_3', xf_3'-zf_1',yf_1'-xf_2')\neq \vec 0 \end{align*} The problem is that others got that it is $\vec 0$. I'm not sure whether who's wrong.