Let $f_X(x_i)=\theta\cdot x_i$, $x_i\leq \sqrt{2/\theta}$ with $\theta=\frac{2}{x_{(n)}^2}$ (derived using the MLE-method). What is the expectation of estimator $\hat\theta$?
I'd assume that $\hat\theta=\frac{2}{\bar{X}_n^2}$, with $\bar{X}_n=\frac{1}{n}\sum_{i=1}^n X_i$
$$\implies E[\hat\theta]=E[\frac{2}{\bar{X}_n^2}]=2E[\frac{1}{(\frac{1}{n}\sum_{i=1}^n X_i)^2}]=2n^2\sum_{i=1}^n E[X_i^{-2}]$$
I however thought this value should be 'unbiased', which is not the case. Recall that 'unbiased' implies that $E[\hat\theta-\theta]\iff E[\hat\theta]=\theta$, which is not true. Could anyone give me a hint where I made a mistake?
As Stefano noted in a comment, $x_{(n)}$ is not the average but the $n$-th order statistic, which for a sample of size $n$ is the maximum. There's a typo in the question; instead of $\theta=\frac{2}{x_{(n)}^2}$ it should say $\hat\theta=\frac{2}{x_{(n)}^2}$; perhaps this is what confused you.
You can tell without calculation that this estimator is biased, since it's always greater than $\theta$.
The cumulative distribution function of the maximum is the $n$-th power of the cumulative distribution function of the individual samples, which is
$$ F(x)=\int_0^x\theta t\,\mathrm dt=\frac12\theta x^2\;, $$
so the expectation of $\hat\theta$ is $$ \int_0^\sqrt{\frac2\theta}\frac2{x^2}\frac{\mathrm d}{\mathrm dx}\left(F(x)^n\right)\mathrm dx=\frac{4n\theta^n}{2^n}\int_0^\sqrt{\frac2\theta}x^{2n-3}\mathrm dx=\frac n{n-1}\theta\;. $$