Use of Poincare Lemma in solving $\nabla \times \textbf{A}(\textbf{r})=\frac{\textbf{r}}{r^3}$ UPDATED 
You are given the following statement of the Poincaré Lemma: If $\Phi_t$ is a one-parameter family of diffeomorphisms on $\mathbb R^n$ (not necessarily a subgroup) and $X_t$ the vector field defined by
  $$X_t \circ \Phi_t = \frac{\partial}{\partial t}\,\Phi_t,$$
  and if $\beta$ is a closed $k$-form on $\mathbb R^n$ such that
  $$\Phi_1^* \beta = \beta, \quad \lim_{\epsilon \to 0} \Phi_\epsilon^* \beta = 0,$$
  then $\beta = d\alpha$, where
  $$\alpha=\int_0^1 \Phi_t^* i_{X_t}\beta\,dt.$$

Let $U = \mathbb{R}^3 \setminus \{(0,0,z) \}$ (i.e. $\mathbb{R}^3$ with the $z$-axis removed ) and consider $\beta$ on $U$ given by
$$\beta = \frac{x \,dy \wedge dz + y \,dz \wedge dx + z \,dx \wedge dy}{(x^2+y^2+z^2)^{3/2}}$$
One can show that $d\beta=0$.
Let $\Phi_t(x,y,z)=(x,y,tz)$. One can show that $\Phi_1^*\beta=\beta$, $\lim_{\epsilon \rightarrow 0} \Phi_{\epsilon}^*\beta=0$.
So use the Poincaré Lemma to find a vector field $\textbf{A}(\textbf{r})$ on $U$ such that
$$\nabla \times \textbf{A}(\textbf{r})=\frac{\textbf{r}}{r^3}$$

I believe that you just need to compute $\displaystyle \alpha=\int_0^1 \Phi_t^* i_{X_t}\beta \,dt$ as $\beta = d\alpha$.
$$
\begin{align}
\hat{\mathbb{X}}_t
&= \left(\frac{\partial}{\partial t}\hat{\Phi}_t \right) \hat{\Phi}_t^{-1} \\
&= \left(\frac{\partial}{\partial t}\hat{\Phi}_t\right) \left(x,y,\frac{z}{t}\right) \\
&=\left(0,0,z/t\right)
\end{align}$$
Now $\Phi_t \,dx =dx$, $\Phi_t\, dy =dy$ and $\Phi_t\, dz = tdz$.
So
$$
\begin{align}
i_{\hat{X}_t}\beta
&= \frac{x}{r^3}i_{\hat{X}_t}(dy \wedge dz)+\frac{y}{r^3}i_{\hat{X}_t}(dz \wedge dx)+\frac{z}{r^3}i_{\hat{X}_t}(dx \wedge dy) \\ 
&= \frac{x}{r^3}\left(\frac{-z}{t}dy\right)+\frac{y}{r^3}\left(\frac{z}{t}dx\right) \\ 
&= \frac{-zxt}{r^3}dy+\frac{zty}{r^3}dx
\end{align}
$$
So
$$
\begin{align}\Phi_t^*i_{\hat{X}_t}\beta
&= \Phi_t^* \left[\frac{-zxt}{r^3}dy+\frac{zty}{r^3}dx \right] \\ 
&=  \frac{-(tz)xt}{(x^2+y^2+t^2z^2)^{3/2}}dy+\frac{(tz)ty}{(x^2+y^2+t^2z^2)^{3/2}}dx \\
&= \frac{-xzt^2}{(x^2+y^2+t^2z^2)^{3/2}}dy+\frac{yzt^2}{(x^2+y^2+t^2z^2)^{3/2}}dx
\end{align}
$$
I am not sure that this is correct as it doesnt look very integrable.
 A: There's a mistake in the last line in the calculation of $i_{\hat{X}_t}\beta$. The $t$'s in the numerators should be in the denominators:
$$i_{\hat{X}_t}\beta = \frac{-zx}{\color{red}{t}r^3}\, dy + \frac{zy}{\color{red}{t}r^3}\, dx = \frac{z}{\color{red}{t}r^3}(y\, dx - x\, dy).$$
Then
$$\Phi_t^*i_{\hat{X}_t}\beta = \frac{z}{(x^2 + y^2 + t^2 z^2)^{3/2}}(y\, dx - x\, dy).$$
Consequently, $$\alpha = I(x,y,z)(y\, dx - x\, dy),$$
where 
$$I(x,y,z) = \int_0^1 \frac{z\, dt}{(x^2 + y^2 + t^2 z^2)^{3/2}}.$$
To evaluate $I(x,y,z)$, let $tz = \sqrt{x^2 + y^2}\tan\theta$. Then $z\, dt = \sqrt{x^2 + y^2}\sec^2\theta\, d\theta$, and thus
\begin{align}I(x,y,z) &= \int_0^{\tan^{-1}\left(\frac{z}{\sqrt{x^2+y^2}}\right)} \frac{\sqrt{x^2+y^2}\sec^2\theta\, d\theta}{(x^2 + y^2)\sqrt{x^2 + y^2} \sec^3\theta}\\
&= \frac{1}{x^2 + y^2} \int_0^{\tan^{-1}\left(\frac{z}{\sqrt{x^2+y^2}}\right)} \cos\theta\, d\theta\\
&= \frac{1}{x^2+y^2}\sin\left(\tan^{-1}\left(\frac{z}{\sqrt{x^2+y^2}}\right)\right)\\
&= \frac{z}{r(x^2+y^2)}.\end{align}
Therefore,
$$\alpha = \frac{yz}{r(x^2 + y^2)}\, dx - \frac{xz}{r(x^2 + y^2)}\, dy$$
and 
$$\mathbf{A}(\mathbf{r}) = \left\langle \frac{yz}{r(x^2 + y^2)}, -\frac{xz}{r(x^2 + y^2)}, 0\right\rangle.$$
