From ##t = \alpha \sinh(\tau/\alpha)##, we have ##z = (t^2 + \alpha^2)^{1/2} = \alpha \cosh(\tau/\alpha)##. Use these to express the denominator of (3.59) in terms of ##\tau## and ##\tau'##. Note ##\mathbf x = z## and ##\mathbf x' = z'##.

Using identities for the hyperbolic sine and cosine functions, show that the denominator of (3.59) may be written to first order in ##\varepsilon## as $$16\pi^2\alpha^2 \left[\sinh^2\left(\frac{\tau-\tau'}{2\alpha}\right) - \frac{i\varepsilon \Delta t}{2\alpha^2}\right].$$ In the last term, ##\Delta t = t - t'##, which is a function of ##\tau## and ##\tau'## that will ultimately be absorbed into ##\varepsilon##.

Once you get this far, you can think about how to get the final result (3.64).