Deriving confidence interval for scale parameter of Weibull distribution 
Let $X_1, \ldots, X_n$ be random variables such that $X_i$ has Weibull distribution with shape parameter $3$ and scale parameter $\theta$ and define $X_{(1)}=\min\{X_1, \ldots ,X_n\}$. Derive $95 \%$ confidence interval for $\theta$ based on $X_{(1)}$.

I know that, but not sure where to go with it : 
The PDF of Weibull distribution is
$$
f(x \mid k,\theta) =
\begin{cases}
k \, \theta \left(x \, \theta \right)^{k-1}e^{-(x \, \theta)^{k}} & x\geq0 ,\\
0 & x<0,
\end{cases}
$$
for $\theta >0$, $k > 0$ is said to have a Weibull distribution, denoted $WEI( k , \theta)$; $k$
and $\theta$ are called its shape and scale parameters, respectively.
The CDF of Weibull distribution is
$$F_{X}(x) = 1- e^{-\left(x \theta \right)^{k}}$$
Now the distribution of $X_{(1)}$ : 
$$F_{X_{(1)}}(x) = \min\{\,X_1,\ldots,X_n\,\} = 1- [ 1 - F_{X}(x) ]^{n}$$
However, I am not really sure how to approach this question. I would really appreciate it if someone could give me some direction. 
Edit 
Attempts according to @NCh 
$$
f(x \mid\theta) =
\begin{cases}
3 \, \theta \left(x \, \theta \right)^{2}e^{-(x \, \theta)^{3}} & x\geq0 ,\\
0 & x<0,
\end{cases}
$$
The CDF of Weibull distribution is
$$F_{X}(x) = 1- e^{-\left(x \theta \right)^{3}}$$
Now, let $W= \theta \, X_{(1)}$
$$
\begin{align}
F_{W}(w) &= P \left( W \le w \right) \\
&= P(X_{(1)} \le \frac{w}{\theta}) \\
&= 1- \left[ 1 - F_{X}(\frac{w}{\theta}) \right]^{n} \\
&= 1- \left[e^{-w^3} \right]^{n}
\end{align}
$$
What I am supposed to do with the $n$. 
Now, 
$$
\begin{align}
P\left(a \le W \le b \right) = P(W \le b) - P(W \le a)=0.95
\end{align}
$$
Where to go from here. Thank you 
 A: Assuming $X_1,X_2,\ldots,X_n$ are independent with common distribution $\mathsf{Weibull}(3,\theta)$.
You have correctly shown that CDF of $W=\theta X_{(1)}$ is $P(W\le w)=1-e^{-n w^3}$ for $w\ge 0$.
So we can take $W$ as a pivot for deriving a confidence interval for $\theta$. 
Suppose that for some $k\,(>0)$,
$$P_{\theta}(W\ge k)=P_{\theta}\left(\theta\ge \frac k{X_{(1)}}\right)=0.95\quad,\,\forall\,\theta>0$$
That is, $k$ is so chosen that $$e^{-nk^3}=0.95 $$
This gives $k^3=-\frac1n\ln(0.95)$, or $k\approx\frac{0.371553}{n^{1/3}}$ (taking the real solution).
So a (one-sided) $95\%$ confidence interval for $\theta$ is simply $\left[\frac k{X_{(1)}},\infty\right)\approx\left[\frac{0.371553}{n^{1/3}X_{(1)}},\infty\right)$.

If you want a two-sided confidence interval, you can use the pivot $2nW^3\sim \chi^2_2$. (Note that $W^3$ has an exponential distribution with mean $1/n$.)
Suppose $\chi^2_{\alpha,2}$ is the upper $100\alpha\%$ point of $\chi^2_2$ distribution, i.e. $P(\chi^2_2>\chi^2_{\alpha,2})=\alpha$ for $\alpha\in(0,1)$.
Then 
$$P_{\theta}\left[\chi^2_{1-\alpha/2,2}\le 2nW^3\le\chi^2_{\alpha/2,2}\right]=1-\alpha\quad,\forall\,\theta>0$$
In your case, this reduces to
$$P_{\theta}\left[\left(\frac{\chi^2_{0.975,2}}{2n X_{(1)}^3}\right)^{1/3}\le \theta\le \left(\frac{\chi^2_{0.025,2}}{2n X_{(1)}^3}\right)^{1/3}\right]=0.95\quad,\forall\,\theta$$
