Vector analysis - Curl of vector 
How to prove it? I have tried several times to solve it, but I still get stuck everytime. 

First Method (long)
Observe that $\operatorname{curl}(\pmb{a}\times\pmb{b})=\nabla\times (\pmb{a}\times\pmb{b})$ where $\nabla=\pmb{i}\frac{\partial}{\partial x}+\pmb{j}\frac{\partial}{\partial y}+\pmb{k}\frac{\partial}{\partial z}$.
Recalling that $\pmb{a}\times\pmb{b}=\left|\matrix{\pmb i & \pmb j & \pmb k\\
a_x & a_y & a_z\\
b_x & b_y & b_z}\right|=(a_yb_z-a_zb_y)\pmb{i}+(a_zb_x-a_xb_z)\pmb{j}+(a_xb_y-a_yb_x)\pmb{k}
$
we have
$$
\operatorname{curl}(\pmb{a}\times\pmb{b})=\nabla\times (\pmb{a}\times\pmb{b})=\left|\matrix{\pmb i & \pmb j & \pmb k\\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\
a_yb_z-a_zb_y & a_zb_x-a_xb_z & a_xb_y-a_yb_x}\right|
$$
The $x$ component is 
$$\small\begin{align}
[\nabla\times (\pmb{a}\times\pmb{b})]_x &=\frac{\partial}{\partial y}(a_yb_z-a_zb_y)-\frac{\partial}{\partial z}(a_xb_z-a_zb_x)=\\
&=a_x\left(\frac{\partial b_y}{\partial y}+\frac{\partial b_z}{\partial z}\right)-b_x\left(\frac{\partial a_y}{\partial y}+\frac{\partial a_z}{\partial z}\right)+
\left(b_y\frac{\partial }{\partial y}+b_z\frac{\partial }{\partial z}\right)a_x-
\left(a_y\frac{\partial }{\partial y}+a_z\frac{\partial }{\partial z}\right)b_x
\end{align}
$$
Adding $a_x\frac{\partial b_x}{\partial x}$ to the first of these four terms, and subtracting from the last, and doing the same with $b_x\frac{\partial a_x}{\partial x}$ to the other two terms, we find
$$
[\nabla\times (\pmb{a}\times\pmb{b})]_x=(\nabla\cdot\pmb{b})a_x-(\nabla\cdot\pmb{a})b_x+(\pmb{b}\cdot\nabla)a_x-(\pmb{a}\cdot\nabla)b_x
$$
and repeating for the $y$ and $z$ components, we find that
$$
\nabla\times (\pmb{a}\times\pmb{b})=(\nabla\cdot\pmb{b})\pmb{a}-(\nabla\cdot\pmb{a})\pmb{b}+(\pmb{b}\cdot\nabla)\pmb{a}-(\pmb{a}\cdot\nabla)\pmb{b}
$$
where the operator $\pmb{a}\cdot\nabla$ is defined as $\pmb{a}\cdot\nabla=a_x\frac{\partial }{\partial x}+a_y\frac{\partial }{\partial y}+a_z\frac{\partial }{\partial z}$ and $\nabla\cdot \pmb{a}=\operatorname{div}\pmb a$.
Now for $\pmb{a}=\pmb{v}$ and $\pmb{b}=\frac{\pmb{r}}{r^3}$ we obtain
$$
\nabla\times \left(\pmb{v}\times\frac{\pmb{r}}{r^3}\right)=
\left(\nabla\cdot\frac{\pmb{r}}{r^3}\right)\pmb{v}-(\nabla\cdot\pmb{v})\frac{\pmb{r}}{r^3}+\left(\frac{\pmb{r}}{r^3}\cdot\nabla\right)\pmb{v}-(\pmb{v}\cdot\nabla)\frac{\pmb{r}}{r^3}
$$
The second and third terms are null, because $\pmb v$ is a constant vector. The first term is the $\operatorname{div}\left(\frac{\pmb{r}}{r^3}\right)$ and is null; infact
$$
\frac{\partial }{\partial x}\left(\frac{x}{\sqrt{x^2+y^2+z^2}}\right)=\frac{1}{r^3}\left(1-\frac{3x^2}{r^2}\right)
$$
Adding this to similar terms for $y$ and $z$ gives
$$
\nabla\cdot\frac{\pmb{r}}{r^3}=\frac{1}{r^3}\left(3-\frac{3(x^2+y^2+z^2)}{r^2}\right)=\frac{1}{r^3}\left(3-\frac{3r^2}{r^2}\right)=0
$$
So we have
$$
\nabla\times \left(\pmb{v}\times\frac{\pmb{r}}{r^3}\right)=-(\pmb{v}\cdot\nabla)\frac{\pmb{r}}{r^3}.
$$
The $x$ component is 
$$\begin{align}
\left[(\pmb{v}\cdot\nabla)\frac{\pmb{r}}{r^3}\right]_x &=\left(a\frac{\partial }{\partial x}+b\frac{\partial }{\partial y}+c\frac{\partial }{\partial z}\right)\frac{x}{\sqrt{x^2+y^2+z^2}}\\
&=a\left(1-\frac{3x^2}{r^2}\right)\frac{1}{r^3}-b\frac{3xy}{r^5}-c\frac{3xz}{r^5}
\end{align}
$$
In the same way, the other components are
$$\begin{align}
\left[(\pmb{v}\cdot\nabla)\frac{\pmb{r}}{r^3}\right]_y &=
&=-a\frac{3xy}{r^5}+b\left(1-\frac{3y^2}{r^2}\right)\frac{1}{r^3}-c\frac{3yz}{r^5}\\
\left[(\pmb{v}\cdot\nabla)\frac{\pmb{r}}{r^3}\right]_z &=
&=-a\frac{3xz}{r^5}-b\frac{3yz}{r^5}+a\left(1-\frac{3z^2}{r^2}\right)\frac{1}{r^3}
\end{align}
$$
It's easy to see that summing up the terms with $-\frac{3}{r^5}$
$$\small
-\frac{3}{r^5}\left[(by+cz)x\pmb{i}+(ax+cz)y\pmb{j}+(ax+by)z\pmb{k}\right]-\frac{3}{r^5}\left[ax^2\pmb{i}+by^2\pmb{j}+cz^2\pmb{k}\right]=
-\frac{3}{r^5}\left[(\underbrace{ax+by+cz}_{\pmb{v}\cdot\pmb{r}})x\pmb{i}-ax^2\pmb{i}+(ax+by+cz)y\pmb{j}-by^2\pmb{j}+(ax+by+cz)z\pmb{k}-cz^2\pmb{k}\right]-\frac{3}{r^5}\left[ax^2\pmb{i}+by^2\pmb{j}+cz^2\pmb{k}\right]=
-\frac{3}{r^5}\left[(\pmb{v}\cdot\pmb{r})(\underbrace{x\pmb{i}+y\pmb{j}+z\pmb{k}}_{\pmb{r}})-ax^2\pmb{i}-by^2\pmb{j}-cz^2\pmb{k}\right]-\frac{3}{r^5}\left[ax^2\pmb{i}+by^2\pmb{j}+cz^2\pmb{k}\right]=-\frac{3}{r^5}(\pmb{v}\cdot\pmb{r})\pmb{r}+\frac{3}{r^5}\left[ax^2\pmb{i}+by^2\pmb{j}+cz^2\pmb{k}\right]-\frac{3}{r^5}\left[ax^2\pmb{i}+by^2\pmb{j}+cz^2\pmb{k}\right]=\color{blue}{-\frac{3}{r^5}(\pmb{v}\cdot\pmb{r})\pmb{r}}
$$
Summing up the terms with $\frac{1}{r^3}$ we find
$$
\frac{a}{r^3}\pmb{i}+\frac{b}{r^3}\pmb{j}+\frac{c}{r^3}\pmb{k}=\frac{1}{r^3}(a\pmb{i}+b\pmb{j}+c\pmb{k})=\color{blue}{\frac{1}{r^3}\pmb{v}}
$$
Putting all together, we have
$$\color{blue}{
\operatorname{curl}\left(\pmb{v}\times\frac{\pmb{r}}{r^3}\right)=\nabla\times \left(\pmb{v}\times\frac{\pmb{r}}{r^3}\right)=-(\pmb{v}\cdot\nabla)\frac{\pmb{r}}{r^3}=\frac{3}{r^5}(\pmb{v}\cdot\pmb{r})\pmb{r}-\frac{1}{r^3}\pmb{v}}
$$
Second Method (shorter)
$$
\begin{align}
[\nabla\times(\pmb{a}\times\pmb{b})]_i 
&=\epsilon_{ijk}\partial_j(\pmb{a}\times\pmb{b})_k\\
&=\epsilon_{ijk}(\partial_j\epsilon_{klm}a_lb_m)\\
&=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})\partial_j(a_lb_m)\\
&=\partial_m(a_ib_m)-\partial_l(a_lb_i)\\
&=(\partial_m a_i)b_m+a_i(\partial_m b_m)-(\partial_l a_l)b_i-a_l(\partial_i b_l)\\
&=(\pmb{b}\cdot\nabla)b_i+(\nabla\cdot\pmb{b})a_i-(\nabla\cdot\pmb{a})b_i-(\pmb{a}\cdot\nabla)b_i
\end{align}$$
so we have finally
$$
\nabla\times (\pmb{a}\times\pmb{b})=(\nabla\cdot\pmb{b})\pmb{a}-(\nabla\cdot\pmb{a})\pmb{b}+(\pmb{b}\cdot\nabla)\pmb{a}-(\pmb{a}\cdot\nabla)\pmb{b}
$$
Now for $\pmb{a}=\pmb{v}$ and $\pmb{b}=\frac{\pmb{r}}{r^3}$ we obtain
$$
\nabla\times \left(\pmb{v}\times\frac{\pmb{r}}{r^3}\right)=
\left(\nabla\cdot\frac{\pmb{r}}{r^3}\right)\pmb{v}-(\nabla\cdot\pmb{v})\frac{\pmb{r}}{r^3}+\left(\frac{\pmb{r}}{r^3}\cdot\nabla\right)\pmb{v}-(\pmb{v}\cdot\nabla)\frac{\pmb{r}}{r^3}
$$
The second and third terms are null, because $\pmb v$ is a constant vector. The first term is the $\operatorname{div}\left(\frac{\pmb{r}}{r^3}\right)$:
$$
\nabla\cdot\frac{\pmb{r}}{r^3}=\partial_i\frac{x_i}{r^3}=\left(1-x_i^2\right)\frac{1}{r^3}\delta_{ii}=0
$$
So we have
$$
\nabla\times \left(\pmb{v}\times\frac{\pmb{r}}{r^3}\right)=-(\pmb{v}\cdot\nabla)\frac{\pmb{r}}{r^3}.
$$
Observing that
$$
(\pmb{v}\cdot\nabla)\frac{\pmb{r}}{r^3}=v_i\partial_i\frac{x_j}{r^3}=\frac{v_i\delta_{ij}}{r^3}-v_ix_jx_i\frac{3}{r^3}=\frac{v_i}{r^3}-(v_ix_i)x_j\frac{3}{r^5}=\frac{\pmb{r}}{r^3}-(\pmb{v}\cdot\pmb{r})\pmb{r}\frac{3}{r^5}
$$
we have finally
$$\color{red}{
\operatorname{curl}\left(\pmb{v}\times\frac{\pmb{r}}{r^3}\right)=\nabla\times \left(\pmb{v}\times\frac{\pmb{r}}{r^3}\right)=-(\pmb{v}\cdot\nabla)\frac{\pmb{v}}{r^3}=\frac{3}{r^5}(\pmb{v}\cdot\pmb{r})\pmb{r}-\frac{1}{r^3}\pmb{v}}
$$
