We have $X_1$ and $X_2$, where $X_i \sim \mathrm{Beta}(\theta,1)$, $f_{X_i}=\theta X^{\theta -1}$ with $0<x<1$.
$H_o: \theta =1$ vs $H_a = \theta = 2$
The rejection region is $C = \{ (x1,x2) \in (0,1)\times (0,1) ; x_1 \ x_2 \geq \frac{3}{4} \}$.
I need to proof that $P( \text{error type 1}) = \alpha = \frac{1}{4} + \frac{3}{4} ln( \frac{3}{4})$.
My first idea is to use the power function $\alpha = \Pi_{\gamma}(\theta_o)= P(x_1 x_2 \geq \frac{3}{4} \ | \ \theta = \theta_1 = 1)$, and I need to know the distribution of $X_1 X_2$, I have found that is complicated to know that distribution.
It is not difficult to compute the desired probability. Under the null hypothesis, $X_i \mid H_0 \sim \operatorname{Beta}(1,1)$ is uniformly distributed; thus the probability that the product $X_1 X_2$ is at least $3/4$ is equal to the area in the unit square above the hyperbola $x_1 x_2 \ge 3/4$; i.e., $$\begin{align} \Pr[X_1 X_2 \ge \tfrac{3}{4} \mid \theta = 1] &= \Pr\left[(3/4 \le X_1 \le 1) \cap \left(\frac{3/4}{X_1} \le X_2 \le 1\right) \Bigl| \, \theta = 1\right] \\ &= \int_{x_1 = 3/4}^1 \!\!\!\Pr\left[ \frac{3/4}{x_1} \le X_2 \le 1 \, \Bigl| \, (X_1 = x_1) \cap (\theta = 1) \right] f_{X_1 \mid \theta = 1}(x_1) \, dx_1 \\ &= \int_{x_1 = 3/4}^1 \left(1 - \frac{3/4}{x_1}\right) \, dx_1 \\ &= \left[x_1 - \frac{3}{4} \log x_1 \right]_{x_1 = 3/4}^1 \\ &= \frac{1}{4} \left( 1 - 3 \log \frac{4}{3} \right) \\ &\approx 0.034238. \end{align}$$