To find critical region we can use Lemma-Pearson Lemma
$$\frac{L_1}{L_0} > k$$
$$\frac{ \prod_{i=1}^{n} \theta (1-x)^{\theta-1}}{ 1*(1-x)^{1-1}} > k,\space \space\space \theta >1$$
$$\prod_{i=1}^{n} \theta (1-x)^{\theta-1} > k,\space \space \space \theta >1$$
$$(\theta)^n \prod_{i=1}^{n}(1-x)^{\theta-1} > k$$
$$\prod_{i=1}^{n}(1-x)^{\theta-1} >\frac{k}{\theta^n}$$
Taking log on both side
$$\log_e( \prod_{i=1}^{n}(1-x)^{\theta-1} )> \log_e(\frac{k}{\theta^n})$$
$$\sum_{i=1}^{n} \log_e ((1-x)^{\theta-1}) > \log_e(\frac{k}{\theta^n})$$
Is my approach correct if correct how do we proceed further in this question?
First observe that, given that $H_1$ is a composite hypothesis actually you have to use another method to find UMP test: as an example, monotone likelikood ratio or Karlin Rubin
After some calculations you find that the Statistic Test is
$$T=\sum_i\log(1-X_i)$$
To find the distribution of $T$, first observe that
$$Y=-\log(1-X)\sim \exp(\theta)$$
and thus $\sum_y Y\sim\text{Gamma}[n;\theta]$ that, under $H_0$ it is a $\text{Gamma}[n;1]$
Concluding:
$$-2T\sim\chi_{(2n)}^2$$
Concluding, the rejection region is
$$P(T>k|\theta=1)=\alpha$$
$$P(-2T<k^*|\theta=1)=\alpha$$
or equivalently
$$-\sum_i \log(1-X_i)^2<\chi_{(2n),1-\alpha}^2$$
This is result A