Take a finite field $\mathbb{F}_{\!q}$ such that $q \equiv 1 \pmod 3$, i.e., $\omega \mathrel{:=} \sqrt[3]{1} \in \mathbb{F}_{\!q}$, $\omega \neq 1$. Also, for $i \in \{0,1,2\}$ consider the elliptic curves $E_i\!: y^2_i = b^ix_i^3 - b$, where $b \in \mathbb{F}_{\!q}^* \setminus (\mathbb{F}_{\!q}^*)^3$. There is on $E_i$ the order $3$ automorphism $[\omega]\!: (x_i,y_i) \mapsto (\omega x_i, y_i)$.

Look at the quotient $T \mathrel{:=} (E_0 \!\times\! E_1 \!\times\! E_2)/[\omega]^{\times 3}$, which is a Calabi–Yau threefold according to Oguiso and Truong - Explicit examples of rational and Calabi–Yau threefolds with primitive automorphisms of positive entropy. It is easily seen that it has the affine model $$ T\!: \begin{cases} y_1^2 + b = b(y_0^2 + b)t_1^3, \\ y_2^2 + b = b^2(y_0^2 + b)t_2^3 \end{cases} \quad \subset \quad \mathbb{A}^{\!5}_{(y_0,y_1,y_2,t_1,t_2)}, $$ where $t_1 \mathrel{:=} x_1/x_0$, $t_2 \mathrel{:=} x_2/x_0$.

Although $T$ is a quite classical quotient, I cannot find a rational $\mathbb{F}_{\!q}$-curve on it. In my opinion, this is a sufficiently interesting algebraic geometry task. Can you help me please? I can explain the origin of this task if it is necessary.