Let $\theta > 0$ a parameter, $Y_1, \dots, Y_n$ is a set of i.i.d. observations with marginal distribution function $F_\theta(y) = [F(y)]^{\dfrac{1}{\theta}} , -\infty < y < +\infty$.
Show that $t = -\dfrac{\sum_{k = 1}^n\log F(y_k)}{n} $ is an MVUE of $ \theta $
I tried to prove that $t$ is unbiased. For that, I had to solve the integral $\int_{-\infty}^{+\infty} \log[F(y_k)] F(y_k)^{\frac{1 - \theta}{\theta}} f(y_k)$. However, I was unable to solve this.
After that, I tried to use the Rao-Blackwell theorem with the sufficient statistic $T = \prod_{k = 1}^n Y_k$ to find that the estimator $E(t \mid T) = t$ and conclude that it is the MVUE. However, I was also unable to do this.
Would someone help me by solving this integral or explaining to me how to do this with the Rao-Blackwell theorem?
Indeed you have to show that $E(T)=\theta$. You are on the right track, but from the Lehmann-Scheffé theorem, if $T$ if a complete sufficient statistic for $\theta$, and $E(g(T))=\tau(\theta)$, then $g(T)$ is the unique MVUE of $\tau(\theta)$.
So, it's important to verify the condition that $T$ is a complete sufficient statistic first. You can do that if you show that $Y$ belongs to the regular exponential family, i.e.
$$ f_{\eta}(y)=b(\eta)h(y)\exp\left\{ \sum_{j=1}^k \eta_j T_j(y)\right\},$$ then $T(Y)=\left(\sum_{i=1}^n T_1(Y_i),\dotsc, \sum_{i=1}^n T_k(Y_i)\right)$, the natural sufficient statistic, is a complete sufficient statistic. (You may want to check http://lagrange.math.siu.edu/Olive/ich4.pdf p. 117)
Moreover, since they give you the marginal distribution, the marginal density can be found: $$ f_{\theta}(y)=\frac{d}{d y}F_{\theta}(y)$$
From all this, showing complete sufficiency should be straightforward (hopefully).
Now, for $E(T)$: when you get this integral (to simplify the notation, I'll put $y$ instead of $y_k$ since it's iid anyway) $$ \int_{-\infty}^{\infty} \log \left(F(y)\right)\left[F(y)\right]^{\frac{1-\theta}{\theta}}f(y)\, dy=\int_{-\infty}^{\infty} \log \left(F(y)\right)\frac{\left[F(y)\right]^{\frac{1}{\theta}}}{F(y)}f(y)\, dy$$ Now try the substitution: $$ u = \log (F(y)) \implies e^u=F(y)\\ du = \frac{1}{F(y)}\cdot F'(y)\, dy$$ but $F'(y)=f(y)$ (don't forget to change your integral bounds). The integral should then be easily done with integration by parts.