Show that the inverse square field \begin{align*} \frac{-X}{\| x\|^3}=-\frac{(x_1, x_2, x_3)}{\left(x_1^2+x_2^2+x_3^2 \right)^{\frac{3}{2}}} \end{align*} has a vector potential \begin{align*} \frac{x_3}{\left( x_1^2+x_2^2+x_3^2 \right)^{\frac{1}{2}}}\frac{\left(-x_2, x_1, 0 \right)}{x_1^2+x_2^2} \end{align*} For all points except along the $x_3$ axis.
A vector field $F$ has a vector potential $G$ when it can be expressed as $F=curl\:G$. So, I have to show:
\begin{align*} curl\left(\frac{x_3}{\left( x_1^2+x_2^2+x_3^2 \right)^{\frac{1}{2}}}\frac{\left(-x_2, x_1, 0 \right)}{x_1^2+x_2^2}\right)=-\frac{(x_1, x_2, x_3)}{\left(x_1^2+x_2^2+x_3^2 \right)^{\frac{3}{2}}} \end{align*}
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