How to prove the following, $$\nabla\left(\frac{\vec{r}}{r}.\vec{r'}\right)=-\frac{\vec{r}}{r}\times\left(\frac{\vec{r}}{r}\times\frac{\vec{r'}}{r}\right)$$ where $\vec{r}$ and $\vec{r'}$ are two different position vectors.
I started the calculation in the following way, $$\nabla(\vec{a}.\vec{b})=(\vec{a}.\nabla)\vec{b}+(\vec{b}.\nabla)\vec{a}+\vec{a}\times(\nabla\times \vec{b})+\vec{b}\times(\nabla\times \vec{a})$$
so, $$\nabla\left(\frac{\vec{r}}{r}.\vec{r'}\right)=\left(\frac{\vec{r}}{r}.\nabla\right)\vec{r'}+(\vec{r'}.\nabla)\frac{\vec{r}}{r}+\frac{\vec{r}}{r}\times(\nabla\times \vec{r'})+\vec{r'}\times\left(\nabla\times \frac{\vec{r}}{r}\right)$$
Now, $$\vec{r'}\times\left(\nabla\times \frac{\vec{r}}{r}\right)=\vec{r'}\times\left(\nabla\left(\frac{1}{r}\right)\times\vec{r}+\frac{1}{r}(\nabla\times\vec{r})\right)=0$$
Also, $$\left(\frac{\vec{r}}{r}.\nabla\right)\vec{r'},\frac{\vec{r}}{r}\times(\nabla\times \vec{r'})=0$$ as $\vec{r'}$ is independent of $\vec{r}$,
I am stuck here, I don't know how to calculate $(\vec{r'}.\nabla)\frac{\vec{r}}{r}$.
I was able to derive the identity. The derivation proceeds as follows,
For a radius vector $\vec{R}=\vec{r}-\vec{r'}$, the magnitude becomces, $$R^2=r^2-2\vec{r}.\vec{r'}+r'^2$$ $$\implies \nabla\left(\frac{R^2}{r}\right)=\nabla(r)-2\nabla(\vec{r}.\vec{r'})+\nabla\left(\frac{r'^2}{r}\right)\\ \implies \frac{2rR\nabla(R)-R^2\nabla(r)}{r^2}=\hat{r}-2\nabla(\vec{r}.\vec{r'})+r'^2\nabla\left(\frac{1}{r}\right)\\ \implies 2\frac{\vec{R}}{r}-R^2\frac{\hat{r}}{r^2}=\frac{\vec{r}}{r}-2\nabla(\vec{r}.\vec{r'})-r'^2\frac{\vec{r}}{r^3}\\ \implies \left(\frac{\vec{R}-\vec{r}}{r}\right)+\frac{\vec{R}}{r}-\frac{\vec{r}}{r^3}(R^2-r'^2)=-2\nabla(\vec{r}.\vec{r'})\\ \implies -\left(\frac{\vec{r'}}{r}\right)+\frac{\vec{R}}{r}-\frac{\vec{r}}{r^3}(r^2-2(\vec{r}.\vec{r'}))=-2\nabla(\vec{r}.\vec{r'})\\ \implies -\left(\frac{\vec{r'}}{r}\right)+\left(\frac{\vec{R}-\vec{r}}{r}\right)+2(\vec{r}.\vec{r'})\frac{\vec{r}}{r^3}=-2\nabla(\vec{r}.\vec{r'})\\ \implies 2\left(\frac{\vec{r'}}{r}\right)-2(\vec{r}.\vec{r'})\frac{\vec{r}}{r^3}=2\nabla(\vec{r}.\vec{r'})$$ Now from the triple product identity of the vector one could write, $$\vec{a}\times(\vec{b}\times\vec{c})=\vec{b}(\vec{a}.\vec{c})-\vec{c}(\vec{a}.\vec{b})$$ So, $$\vec{r}\times(\vec{r}\times\vec{r'})=\vec{r}(\vec{r}.\vec{r'})-\vec{r'}(\vec{r}.\vec{r})\\ \implies \frac{\vec{r}\times(\vec{r}\times\vec{r'})}{r^3}=\frac{\vec{r}}{r^3}(\vec{r}.\vec{r'})-\frac{\vec{r'}}{r}\\ \implies \frac{\vec{r}\times(\vec{r}\times\vec{r'})}{r^3}+\frac{\vec{r'}}{r}=\frac{\vec{r}}{r^3}(\vec{r}.\vec{r'})$$ Putting this back into the above equation gives the proves the identity in the question.