The proof is not exactly standard, although it relates to the "law of the unconscious statistician" [an expression I cannot fathom and do not find amusing] :
First, define $Y=\min\{X,\mu^2/X\}$ which belongs to $(0,\mu)$. The density of $Y$ can be derived from $(y<\mu)$
$$\mathbb P(Y\le y) = \mathbb P(X\le y)+\mathbb P(\mu^2/X \le y\,,\,X>\mu)$$
as
$$f_Y(y;\mu,\lambda)=\left\{f_X(y)+\frac{\mu^2}{y^2}f_X(\mu^2/y)\right\}\mathbb I_{(0,\mu)}(y)$$
And if we notice that
$$\dfrac{(\mu-\mu^2/y)^2}{\mu^2\,\mu^2/y}=\dfrac{(\mu-\mu^2/y)^2}{\mu^2\,\mu^2/y}=\dfrac{(\mu-y)^2}{\mu^2\,y}$$
which is also why $Z=\frac{(X-\mu)^2}{\mu^2X}$, then
\begin{align}f_Y(y;\mu,\lambda)&=\sqrt{\frac{\lambda}{2\pi}}\,e^{-\frac{\lambda(\mu-y)^2}{2\mu^2\,y}}\left\{y^{-3/2}+\mu^{-1}\,y^{-1/2} \right\}\\
&=\sqrt{\frac{\lambda}{2\pi}}\,e^{-\frac{\lambda(\mu-y)^2}{2\mu^2\,y}}\,y^{-3/2}\mu^{-1}\,(\mu+y)\end{align}
If we consider the transform$$H(y) = \dfrac{\lambda(\mu-y)^2}{\mu^2\,y}$$
then
\begin{align}\left\vert\dfrac{\text{d}H(y)}{\text{d}y}\right\vert
&=\frac{\lambda}{\mu^2}
\frac{(\mu-y)}{y}\left\{\frac{\mu-y}{y}+2 \right\}\\
&=\frac{\lambda}{\mu^2}\frac{(\mu-y)(\mu+y)}{y^2}\\
&=\frac{\sqrt{\lambda}}{\mu}H(y)^{1/2}\frac{(\mu+y)}{y^{3/2}}
\end{align}
Which leads to
$$\require{enclose}
f_Y(y;\mu,\lambda)\text{d}y=\frac{1}{\sqrt{2\pi}}\,e^{-z/2}\,z^{-1/2}\frac{\text{d}z}{\enclose{horizontalstrike}{\text{d}y}}\,\enclose{horizontalstrike}{\text{d}y}=f_Z(z;\mu,\lambda)\text{d}z$$
i.e., a chi-square $\chi^2(1)$ density.
Note that a proof of the above using the moment generating function of $Z$ is straightforward (communication of Éric Marchand from Sherbrooke) and that Seshadri's 1994 book The Inverse Gaussian Distribution is the ultimate reference in the matter (communication of Gérard Letac).
