How to prove $\mathcal{I}_{1}(\eta) = \mathcal{I}_{1}(\theta)[h'(\eta)]^{2}$ where $\mathcal{I}_{1}$ is the Fisher information and $\theta = h(\eta)$? I am trying to apply the following definition of Fisher Information:
\begin{align*}
\mathcal{I}_{1}(\theta) = \mathbb{E}_{\theta}\left[\left(\frac{\partial}{\partial\theta}\ln f(x_{1}|\theta)\right)^{2}\right]
\end{align*}
But I have not succeeded so far. This is my attempt so far:
\begin{align*}
\mathcal{I}_{1}(\eta) & = \mathbb{E}_{\eta}\left[\left(\frac{\partial}{\partial\eta}\ln f(x_{1}|\eta)\right)^{2}\right]\\\\
& = \mathbb{E}_{\eta}\left[\left(\frac{\partial\theta}{\partial\eta}\frac{\partial}{\partial\theta}\ln f(x_{1}|\eta)\right)^{2}\right]
\end{align*}
Can someone help me to finish the demonstration?

It's just a consequence of the chain rule. The density of $X$ is $f(x; \theta) = f(x; h(\eta))$. Therefore, viewing it as a function of $\eta$, we have
\begin{align}
\frac{d\ln f(x; h(\eta))}{d\eta} = \frac{1}{f(x; h(\eta))}\frac{df(x; \theta)}{d\theta}\frac{d\theta}{d\eta} = \frac{1}{f(x; \theta)}\frac{df(x; \theta)}{d\theta}h'(\eta) = \frac{d\ln f(x; \theta)}{d\theta}h'(\eta).
\end{align}
It then follows that
\begin{align}
I(\eta) &= \int_{\mathbb{R}}\left[\frac{d\ln f(x; h(\eta))}{d\eta}\right]^2f(x;h(\eta))dx \\
&= \int_{\mathbb{R}}\left[\frac{d\ln f(x; \theta)}{d\theta}h'(\eta)\right]^2f(x; \theta)dx \\
&= [h'(\eta)]^2\int_{\mathbb{R}}\left[\frac{d\ln f(x; \theta)}{d\theta}\right]^2f(x; \theta)dx \\
&= [h'(\eta)]^2I(\theta).  
\end{align}
