"Show that the mean of a random sample of sizenis a minimum variance unbiased estimator of the parameter $\lambda$ of a Poisson distribution."

Here are my steps. Somewhere along the way, I got lost.

1. The distribution is Poisson, so $\mu = \sigma^2 = \lambda$. $E(\overline{X})$, so that the estimator is unbiased is clear.

2. $f(x) = \frac{\lambda ^ x \exp(-\lambda)}{x!}$

3. $\ln f(x) = x \ln \lambda - \ln x! - \lambda$

4. $\{\frac{\partial}{\partial \lambda}[\ln f(x)] \}^2 = \frac{x^2}{\lambda^2} - \frac{2x}{\lambda} + 1$

There's no way that can be right...What did I do wrong??