Proof inverse Gaussian distribution belongs to the exponential family
$$ f(y;\theta,\phi)=\exp\left\{\frac{y\theta-b(\theta)}{a(\phi)}+c(y,\phi)\right\}. $$
What I have gotten so far:
The probability density function of inverse Gaussian distribution is
$$ f(y)=\left(\frac{\lambda}{2\pi y^3}\right)^{\frac{1}{2}}\exp\left\{ -\frac{\lambda}{2\mu^2}\frac{(y-\mu)^2}{y} \right\} $$
where $y\gt0$, $\mu\gt0$, and $\lambda\gt0$ and $Y\sim IG(\mu,\lambda).$
Now I can manipulate the probability density function.
$$ f(y)=\exp\left\{\log\left(\frac{\lambda}{2\pi y^3}\right)^{\frac{1}{2}}\right\}\exp\left\{ -\frac{\lambda}{2\mu^2}\frac{(y-\mu)^2}{y} \right\} \\ = \exp\left\{ \frac{1}{2}\log\lambda-\frac{1}{2}\log2\pi y^3 -\frac{\lambda}{2\mu^2}\frac{(y-\mu)^2}{y} \right\}.$$
I need to reparametrize the parameters $\mu$ and $\lambda$ in respect to $\theta$ and $\phi$ in order to get the probability density function to the form $f(y;\theta,\phi)$. However, I am unsure for to choose these parameters.
Equating two expressions for the log-pdf, $$\color{red}{\frac{y\theta-b(\theta)}{a(\phi)}}\color{blue}{+c(y,\,\phi)}=\color{red}{-\frac{\lambda}{2\mu^2}y+\frac{\lambda}{\mu}}\color{blue}{+\frac12\ln\frac{\lambda}{2\pi y^3}-\frac{\lambda}{2y}}.$$So take$$\phi=\lambda,\,a=\frac{1}{\phi},\,\theta=-\frac{1}{2\mu^2},\,b=-\sqrt{-2\theta},\,c=\frac12\ln\frac{\phi}{2\pi y^3}-\frac{\phi}{2y}.$$