Show $E\left[\left(\frac{\partial}{\partial\theta} \ln f(X)\right)^2\right]=-E\left[\frac{\partial^2}{\partial\theta^2} \ln f(X)\right]$

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I encountered a question given $\theta$ in a random sample of size $n$ and also

$$-n\times E\left[\frac{\partial^2}{\partial \theta^2}\times \ln f(X)\right]$$

, where $f(x)$ is the p.d.f. at $x$, provided that the extremes of the region for which $f(x) \neq0$ do not depend on $\theta$. The derivation of this formula take the following steps:

(a) Differentiating the expressions on both sides of $\int f(x)\,dx=1$ with respect to $\theta$, show that

$$\int \frac{\partial}{\partial\theta} \ln f(x)\times f(x)\,dx=0$$

by interchanging the order of integration and differentiation.

Now I am having trouble understanding what this question was asking me to do. I used

$$\int f(x)\,dx=1$$

and then,

$$\int \frac{\partial}{\partial\theta} f(x)\, dx = 0$$

But did not know how to proceed to get the answer.

In case anyone want to know, the (b) is: Differentiating again with respect to $\theta$, show that.

$$E\left[\left(\frac{\partial}{\partial\theta} \ln f(X)\right)^2\right]=-E\left[\frac{\partial^2}{\partial\theta^2} \ln f(X)\right]$$

Thank you very much for your reading and any assistance would be appreciated!! Thanks!