The potential energy between two dipoles with dipole moments $\mathbf{m_1}$ and $\mathbf{m_2}$ on the distance $r$ can be expressed by $$\phi = (\mathbf{m_1} \cdot \nabla)( \mathbf{m_2} \cdot \nabla) \dfrac{1}{r}.$$
Expand the expression in terms of scalar products.
Attempted solution
Starting at the right side, we obtain:
$( \mathbf{m_2} \cdot \nabla) \dfrac{1}{r} = -\dfrac{1}{r^2}\mathbf{m_2} \cdot\hat{e}_r$.
Going back to the expression, we arrive at: $$\phi = \mathbf{m_1}\cdot \nabla(-\dfrac{1}{r^2}\mathbf{m_2} \cdot\hat{e}_r) = \mathbf{m_1}\cdot [-\dfrac{1}{r^2}\nabla(\mathbf{m_2} \cdot\hat{e}_r)-(\mathbf{m_2} \cdot\hat{e}_r)\nabla(\dfrac{1}{r^2})] = ...?$$
I know that $\nabla(1/r^2) = -\dfrac{2}{r^3}\hat{e}_r$ so it's really the term $\nabla(\mathbf{m_2}\cdot \hat{e}_r)$ that is throwing me off. I know that $\nabla(\mathbf{A} \cdot \mathbf{B}) = (\mathbf{A} \cdot \nabla) \mathbf{B} + (\mathbf{B} \cdot \nabla) \mathbf{B} + \mathbf{A} \times(\nabla \times \mathbf{B}) + \mathbf{B} \times(\nabla \times\mathbf{A)}$ but I'm not sure that is helping. Any advice?
You're almost there. I will take it from $$ \phi = -\frac{1}{r^2}\mathbf{m_1} \cdot \nabla(\mathbf{m_2}\cdot \hat{e}_r) - (\mathbf{m_2}\cdot \hat{e}_r) \left(\mathbf{m_1}\cdot \nabla \left(\frac{1}{r^2} \right)\right). $$
Now, in indices $\mathbf{m_1} \cdot \nabla(\mathbf{m_2}\cdot \hat{e}_r)= (m_1)_j \partial_j [(m_2)_k (\hat{e}_r)_k] $. Since $(m_2)_k=const$, we may pull it out of the derivative and find:
$$ \mathbf{m_1} \cdot \nabla(\mathbf{m_2}\cdot \hat{e}_r) = (\mathbf{m_1} \otimes \mathbf{m_2}) : (\nabla \otimes \hat{e}_r), $$ where $\otimes$ is the tensor product and $:$ is the double dot product. Now, since $\hat{e}_r = \mathbf{r}/r$ with $\mathbf{r}$ position and $r=|\mathbf{r}|$, $$[\nabla \otimes \hat{e}_r]_{ij} = \partial_{i}\left(\frac{r_j}{r}\right) = \frac{\delta_{ij}}{r} - \frac{\mathbf{r}\otimes \mathbf{r}}{r^3}. $$
Hence, we find: $$ \phi = \frac{3(\mathbf{m_1}\cdot \mathbf{r})(\mathbf{m_2}\cdot\mathbf{r})}{r^5} - \frac{\mathbf{m_1}\cdot\mathbf{m_2}}{r^3}. $$