suppose p(y|theta) is poisson distribution with mean parameter theta. show that the sample mean is an efficient estimator theta. 
(Problem 9.3) Suppose $p(y \mid \theta)$ is a Poisson distribution with mean parameter $\theta$.  Show that the sample mean is an efficient estimator of $\theta$.

Work shown:

$$\hat V(\theta) = \frac{1}{n \operatorname{E}[(\frac{\partial}{\partial\theta} \log f(y \mid \theta))^2]}$$
  $$\begin{align*} \frac{\partial}{\partial \theta} \log f(y \mid \theta) &= \frac{\partial}{\partial \theta} \left[\log \frac{e^{-\theta} \theta^y}{y!} \right] = \frac{\partial}{\partial \theta} \left[ \log e^{-\theta}  \theta^y \frac{1}{y!} \right] \\ 
&= \frac{\partial}{\partial \theta} \left[ -\theta + y \log \theta - \log y! \right] \\
&= -1 + \frac{y}{\theta}. \end{align*} $$

I can do only basic calculus, so I need detailed explanation about this problem.
If you can't describe it smoothly, because of the online environment, please let me know a book(or chapter name) that I can refer to.
q1. Does the symbol before the $\log$ have the same meaning as $d$(differential)?
     * it's very good for me, if you illustrate about this little bit...
q2. I don't know how this equation(note) has proceeded.
    Please explain it to me step-by-step.
q3.  $f(y\mid \theta)$ is Bayes' rule?
     I think it isn't, but if isn't why is it written like that? Also, what is it?!
Thanks you for reading despite my poor grammar. 
 A: Q1.  The symbol $\partial$ refers to partial differentiation.  If one has a function of a single variable, for example $$f(x) = x^2,$$ then the first derivative of this function with respect to this single variable would be written as $$\frac{d}{dx}\left[f(x)\right] = \frac{df}{dx} = 2x.$$  If one has a function of more than one variable, e.g., $$f(x,y) = x^2 y + y^3 - 12,$$ then we can compute partial derivatives with respect to either $x$ or $y$, and they are written $$\frac{\partial}{\partial x}\left[f(x,y)\right] = \frac{\partial f}{\partial x} = 2xy, \quad \frac{\partial}{\partial y} \left[f(x,y)\right] = \frac{\partial f}{\partial y} = x^2 + 3y^2.$$  In this case, the reason why we use a partial derivative is because the log-likelihood $\log f(y \mid \theta)$ is a function of $y$ as well as $\theta$.
Q2.  Your question is too broad.  You need to be more specific about what you do not understand about this solution.  Generally, what is being computed here is the Cramer-Rao lower bound for the variance of the sample mean of an IID sample $\boldsymbol y = (y_1, \ldots, y_n)$ of size $n$ drawn from a Poisson distribution with unknown mean $\theta$.  Then, if we can show that the variance of the sample mean of this estimator attains this bound, that this estimator is a UMVUE (since the sample mean is unbiased).
I should also point out here that the notation that was used above is incorrect:  $\hat V (\theta)$ does not make sense.  Instead, it should read $$\operatorname{Var}[\hat\theta],$$ and the equality after this is not actually an equality, but an inequality; i.e., it should read $$\operatorname{Var}[\hat\theta] \ge \frac{1}{n \operatorname{E}[(\frac{\partial}{\partial\theta} \log f(y \mid \theta))^2]}.$$  This is because you are calculating a lower bound on the variance of the estimator $\hat \theta$ which in this case is the sample mean $\bar y = \frac{1}{n} \sum_{i=1}^n y_i$.  You are not calculating an estimate of the variance of a parameter, which is what $\hat V(\theta)$ notates.  A parameter is not a random variable--it has no variance to speak of and no distributional assumption has been applied in this context.
Q3.  $f(y \mid \theta)$ describes the conditional density of a distribution, given the parameter $\theta$.
