Let $X_1, ...,X_n$ independent sample variables of the random variable $X$. Show that the Maximum likelihood estimator $S$ for $\theta := \frac{1}{\lambda}$ is given by:
$$S = \frac{\sum_{i = 1}^n X_i^3}{n}$$
The exponential distribution is defined as:
$$f_X(x) = \lambda \cdot 3x^2 \cdot e^{-\lambda \cdot x^3} \ \ \ \ \ \ \text{for } x \geq 0 $$
I tried everything this, but the derivation is still a problem.
$$(1) \ L (\theta) = \prod_{i = 1}^n \lambda \cdot 3x_i^2 \cdot e^{-\lambda \cdot x_i^3} = e^{-\lambda \cdot x_i^3} \cdot \prod_{i=1}^n \lambda \cdot 3x_i^2$$ $$(2) \ \ \ln(L(\theta)) = -n^3 \cdot \lambda + \sum_{i = 1}^n ln(\lambda) \cdot ln(3x_i)^2$$
$$(3) \ \ ln'(L(\theta)) = - n^3 + \sum_{i = 1}^n \frac{1}{\lambda} \cdot \frac{1}{3x_i^2} = 0$$
$$(4) \ \ \frac{1}{\lambda} \sum_{i = 1}^n \frac{1}{3x_i^2} = n^3$$
$$(5) \ \ \lambda = \frac{n^3}{\sum_{i = 1}^n 3x_i^2}$$
That's don't look like S above.
Would be glad, if someone has a hint.
Your expression rearranging the likelihood is not quite correct.
$$ L (\theta) = \prod_{i = 1}^n \lambda \cdot 3x_i^2 \cdot e^{-\lambda \cdot x_i^3} = \lambda^n \cdot 3^n \cdot e^{-\lambda \cdot \sum x_i^3} \cdot \prod_{i=1}^n x_i^2$$
$$\log_e(L(\theta)) = n \log_e(\lambda) + n \log_e(3) -\lambda \sum x_i^3 + 2 \sum \log_e(x_i)$$
The rest you can probably do yourself