I have an identity
$$\vec{\nabla} \times (\frac{\vec{m} \times \hat{r}}{r^2})$$
which should give us
$$3(\vec{m} \cdot \hat{r}) \hat{r} - \vec{m}$$
But I have to derive it using the Einstein summation notation.
How can I approach this problem to simplify things ?
I can write this but then I am not sure how to proceed
$$∂_w\hat{e}_w \times (\frac{m_ir_j\epsilon_{ijk}\hat{e}_k}{r_l^2})$$
Your $r^2$ should actually be $r^3=(r_l^2)^{3/2}$, because $\hat{r}=r^{-1}\vec{r}$. In a moment, we'll use the $k=-\frac32$ case of$$\partial_wr_l^2=2r_w\implies\partial_w(r_l^2)^k=2kr_w(r_l^2)^{k-1}.$$There's another mistake: dimensional analysis tells us the cross product's units are those of $r^{-3}m$, so the intended result should be divided by $r^3$ too. Given this, the desired result is $\frac{3\vec{m}\cdot\vec{r}}{r^5}\vec{r}-\frac{\vec{m}}{r^3}$. Indeed, use $\hat{e}_w\times\hat{e}_k=\epsilon_{wkm}\hat{e}_m$ to rewrite the cross product as$$\begin{align}\epsilon_{ijk}\epsilon_{wkm}m_i\partial_w\left(\frac{r_j}{r^3}\right)\hat{e}_m&=(\delta_{jw}m_m-\delta_{jm}m_w)\frac{r^2\delta_{jw}-3r_jr_w}{r^5}\hat{e}_m\\&=\frac{2m_mr^2+3r_w(m_wr_m-m_mr_w)}{r^5}\hat{e}_m\\&=\frac{3\vec{m}\cdot\vec{r}}{r^5}\vec{r}-\frac{\vec{m}}{r^3}.\end{align}$$