a) It is easy to show that $d\beta=0$.
b)
$\begin{align}\hat{\mathbb{X}}_t &= \left(\frac{\partial}{\partial t}\hat{\Phi}_t \right) \hat{\Phi}_t^{-1} \\ &= \left(\frac{\partial}{\partial t}\hat{\Phi}_t\right) \left(\frac{x}{t},y,z\right) \\ &=\left(\frac{x}{t},0,0\right) \end{align}$
Now $\Phi_t^* x =tx$, $\Phi_t^* y =y$, $\Phi_t^* z =z$ , so,
$$ \hat{\Phi}_t^* \beta = \hat{\Phi}_t^* \left(\frac{(t^2x^2-y^2)dy \wedge dz + 2txyt dz \wedge dx}{(t^2x^2+y^2)^2}\right) \rightarrow \frac{-1}{y^2} dy \wedge dz $$
So chose $\alpha_0$ to be $(t^2x^2+y^2)^2=\frac{1}{y}dz$
c) First want to compute that $i_{\mathbb{X}_t}\beta$.
$$\begin{align} i_{\mathbb{X}_t} \beta &= i_{\mathbb{X}_t} \left(\frac{(x^2-y^2)}{\rho^4} dy \wedge dz\right) + i_{\mathbb{X}_t}\left(\frac{2xy }{\rho^4} dz \wedge dx\right) \\ &= \left(\frac{(x^2-y^2)}{\rho^4}\right)\left[dy(\mathbb{X}_t)dz - dz(\mathbb{X}_t)dy\right]+\frac{2xy}{\rho^4}[dz(\mathbb{X}_t)dx-dx(\mathbb{X}_t)dz] \\ &= \left(\frac{2xy}{\rho^4}\right)[\frac{-x}{t}dz] \end{align}$$
Then I have that,
$$\begin{align} \Phi_t^*(i_{\mathbb{X}_t}\beta) &= \frac{-2y}{t}\left(\frac{t^2x^2}{(t^2x^2+y^2)^2}\right)dz \\ &= -2yt \left(\frac{x^2}{(t^2x^2+y^2)^2}\right)dz \end{align}$$
But this isnt very integrable. Not sure where I have gone wrong or how I should proceed with this question.
You just made a silly mistake in the last line of computing $i_{\mathbb{X}_t}\beta$ and grabbed the wrong coefficient.
You should have $2xy$ in the numerator of the first factor instead of $x^2-y^2$. Then what you get after the pullback is easily integrated.