A Vector Identitiy with the Levi Civita Symbol I am trying to prove:

$$ \nabla  \times (f \mathbf{r}) = 0  $$

I have done the following:

$$ \nabla  \times (f \mathbf{r}) = \epsilon_{ijk}\partial_j(f \mathbf{r})_k  = \epsilon_{ijk}[(\partial_jf)r_k + f\partial_jr_k], $$

where $f$ is a scalar function.
The second term is zero, and we have by differentiating $f$ via the chain rule:

$$ \epsilon_{ijk} f'(r) (\partial r_j)r_k =  \epsilon_{ijk} f'(r) \frac{r_j}{r} r_k  $$.

If the working is correct, can't seem to show how and why this is equal to zero.
 A: 
I thought it might be instructive to present an approach that uses explicit notation for the Levi-Cevita.  To that end we proceed. 

Using Einstein's summation notation with implied summation over repeated indices, we can write
$$\begin{align}
\left(\nabla \times (f(|\vec r|) \vec r)\right)\cdot \hat x_i&=\hat x_i \cdot \left(\hat x_j\times \hat x_k \partial _j (f(|\vec r|) x_k)\right) \tag 1\\\\
&=\hat x_i \cdot (\hat x_j \times \hat x_k)\left(\delta_{jk}f(|\vec r|+f'(|\vec r|)\frac{x_kx_j}{|\vec r|}) \right) \tag 2\\\\
&=(\hat x_i\cdot \color{blue}{(\hat x_j\times \hat x_j)})f(|\vec r|)+\hat x_i \cdot \color{red}{(\vec r\times \vec r)}\frac{f'(|\vec r|)}{|\vec r|} \tag 3\\\\
&=(\hat x_i \cdot \color{blue}{0})f(|\vec r|)+\hat x_i \cdot \color{red}{0}\frac{f'(|\vec r|)}{|\vec r|} \tag 4\\\\
&=0
\end{align}$$
In going from $(1)$ to $(2)$, we used the product rule for differentiation along with the identity $\frac{\partial x_k}{\partial x_j}=\delta_{jk}$, where $\delta_{jk}$ is the Kronecker Delta.
In going from $(2)$ to $(3)$, we applied the sifting property of the Kronecker Delta to obtain the term in blue-colored font.  In addition, we noted that $\hat x_j x_j=\vec r$ to obtain the term in red-colored font.
In going from $(3)$ to $(4)$, we used the fact that $\vec A\times \vec A=0$ for any vector $\vec A$.
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With
  $\ds{x_{1} \equiv x\,,\ x_{2} \equiv y\,,\ x_{3} \equiv z\,,\
\hat{e}_{1} \equiv \hat{x}\,,\ \hat{e}_{2} \equiv \hat{y}\,,\ \hat{e}_{3} \equiv \hat{z}}$ and 
  sum indexes belong to $\ds{\braces{1,2,3}}$:

\begin{align}
&\color{#f00}{\nabla\times\bracks{\mrm{f}\pars{r}\vec{r}}} =
\sum_{i}\hat{e}_{i}\braces{\nabla\times\bracks{\mrm{f}\pars{r}\vec{r}}}_{i} =
\sum_{i}\hat{e}_{i}\braces{\sum_{jk}\epsilon_{ijk}
\partiald{}{x_{j}}\bracks{\mrm{f}\pars{r}\vec{r}}_{k}}
\\[5mm] = &\
\sum_{i}\hat{e}_{i}\sum_{jk}\epsilon_{ijk}
\partiald{}{x_{j}}\bracks{\mrm{f}\pars{r}x_{k}} =
\sum_{i}\hat{e}_{i}\sum_{jk}\epsilon_{ijk}
\bracks{\partiald{\mrm{f}\pars{r}}{x_{j}}\,x_{k} +
\mrm{f}\pars{r}\,\partiald{x_{k}}{x_{j}}}\label{1}\tag{1}
\end{align}

$$
\mbox{Note that}\
\left\{\begin{array}{rcl}
\ds{\partiald{\mrm{f}\pars{r}}{x_{j}}} & \ds{=} &
\ds{\totald{\mrm{f}\pars{r}}{r}\,\partiald{r}{x_{j}} =
\mrm{f}'\pars{r}\,\pars{{1 \over 2r}\,\partiald{r^{2}}{x_{j}}} = 
\mrm{f}'\pars{r}\pars{{1 \over 2r}\,2x_{j}} = {\mrm{f}'\pars{r} \over r}\,x_{j}}
\\[2mm]
\ds{\partiald{x_{k}}{x_{j}}} & \ds{=} & \ds{\delta_{kj}}
\end{array}\right.
$$

With these results, \eqref{1} is reduced to
\begin{align}
&\color{#f00}{\nabla\times\bracks{\mrm{f}\pars{r}\vec{r}}} =
\sum_{i}\hat{e}_{i}\bracks{{\mrm{f}'\pars{r} \over r}\sum_{jk}\epsilon_{ijk}
x_{j}x_{k} +
\mrm{f}\pars{r}\sum_{jk}\epsilon_{ijk}\,\delta_{kj}}
\\[5mm] = &\
\sum_{i}\hat{e}_{i}\bracks{{\mrm{f}'\pars{r} \over r}\,{1 \over 2}\
\sum_{jk}\underbrace{\pars{\epsilon_{ijk} + \epsilon_{ikj}}}_{\ds{=\ 0}}\
x_{j}x_{k} +
\mrm{f}\pars{r}\sum_{j}\underbrace{\epsilon_{ijj}}_{\ds{=\ 0}}} =
\color{#f00}{\vec{0}}
\end{align}
A: You're almost there.
$$\epsilon_{ijk}r_jr_k \stackrel{(*)}= \epsilon_{ikj}r_kr_j \stackrel{(\dagger)}= -\epsilon_{ijk}r_jr_k  \\ \implies \epsilon_{ijk}r_jr_k = 0$$
$*:$ Renaming dummy indices: $j\leftrightarrow k$.
$\dagger:$ $\epsilon_{ikj}=-\epsilon_{ijk}$ and commuting $r_k$ and $r_j$ (which are just scalars).
