Find the distribution function of $\theta/X_{(1)}$. For $t\in(0,1)$, one should end up with
\begin{align}
P\left[\frac{\theta}{X_{(1)}}\le t\right]&=P\left[X_{(1)}\ge \frac{\theta}{t}\right]
\\&=\left\{P\left[X_1\ge \frac{\theta}{t}\right]\right\}^n
\\&=t^n
\end{align}
Now there exists $(\ell_1,\ell_2)$ with $0\le \ell_1<\ell_2\le 1$ such that $$P_{\theta}\left[\ell_1<\frac{\theta}{X_{(1)}}<\ell_2\right]=P_{\theta}\left[\ell_1X_{(1)}<\theta<\ell_2X_{(1)}\right]=1-\alpha\quad\,\forall\,\theta>0 \tag{1}$$
This gives a confidence interval $(\ell_1X_{(1)},\ell_2X_{(1)})$ for $\theta$. Expected length of this interval is
$$E[\ell_2X_{(1)}-\ell_1X_{(1)}]=(\ell_2-\ell_1)E[X_{(1)}]$$
Since $E[X_{(1)}]$ is a constant, need only to minimize $\ell_2-\ell_1=f$ (say) subject to $(1)$, that is $$\ell_2^n-\ell_1^n=1-\alpha \tag{2}$$
This constrained optimization problem can be solved by usual calculus methods.
Differentiating both sides of $(2)$ with respect to $\ell_2$, we have
$$n\ell_2^{n-1}-n\ell_1^{n-1}\frac{d\ell_1}{d\ell_2}=0$$
Or, $$\frac{d\ell_1}{d\ell_2}=\left(\frac{\ell_2}{\ell_1}\right)^{n-1}$$
Therefore differentiating $f$ we get
$$\frac{df}{d\ell_2}=1-\frac{d\ell_1}{d\ell_2}=1-\left(\frac{\ell_2}{\ell_1}\right)^{n-1}<0$$
Hence $f$ is decreasing in $\ell_2$, so its minimum occurs when $\ell_2$ is maximum.
