Lots of definitions here.
My definition for the fourier transform is $$\widehat{f}(w)=\int_{-\infty}^{\infty}f(t)e^{-iwt}dt$$
$1$) I must show that $$\int_{-\infty}^{\infty}|Sf(\mu, \varepsilon)|^2d\varepsilon=2\pi$$ where $Sf(\mu,\epsilon)$ is the short-time fourier transform with window function $g$ $$Sf(\mu, \varepsilon)=\int_{-\infty}^{\infty}f(t)g(t-\mu)e^{-i\varepsilon t}dt$$ and in this case the function is $f(t)=e^{i\phi(t)}$ where $\phi(t)$ is the father wavelet defined as $$|\widehat{\phi}(w)|^2=\int_{1}^{\infty}|\widehat{\Psi}(sw)|^2\frac{ds}{s}=\int_{|w|}^{\infty}\frac{|\widehat{\Psi}(\varepsilon)|^2}{\varepsilon}d\varepsilon$$ where $\Psi$ is the Mexican hat wavelet defined as the negative second derivative of a Gaussian $$\Psi(t)=\frac{2}{\pi^{1/4}\sqrt{3\sigma}}\Big(1-\frac{t^2}{\sigma^2}\Big)\text{exp}\Big(-\frac{t^2}{2\sigma^2}\Big)$$
$2$) I also must show that $$\int_{-\infty}^{\infty}\varepsilon |Sf(\mu, \varepsilon)|^2d\varepsilon = 2\pi \int_{-\infty}^{\infty}\phi'(t)|g(t-\mu)|^2dt$$
I don't know how to go about solving $2$) but it probably involves $1$). For $1$) The Short time fourier transform is the fourier transform of $f(t)g(t-\mu)$ so by Plancherel's identity I would just need to prove $$\int_{-\infty}^{\infty}|f(t)g(t-\mu)|^2\:dt=1$$ but I don't know how to proceed. Something else that might help are the following, although I would need a proof for the second theorem. $$2\pi \int_{-\infty}^{\infty}|f(t)|^2\:dt=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}|Sf(\mu, \varepsilon)|^2\:d\varepsilon \: d\mu$$ $$2\pi ||\phi||_2^2=||\widehat{\phi}||_2^2=1$$