Consider the Fermat sextic curve $F: x^6 + y^6 + 1 = 0$ over an algebraically closed field of characteristic $0$. It has the two order $3$ automorphisms $\omega_x(x,y) := (\omega x, y)$ and $\omega_y(x,y) := (x, \omega y)$, where $\omega$ is a primitive cubic root of unity. Obviously, the quotient $F/\langle \omega_x, \omega_y \rangle$ is the diagonal conic $x^2 + y^2 + 1$, i.e., this quotient is a rational curve. Further, consider the symmetric square $F^{(2)}$ of the curve $F$. The automorphisms $\omega_x$, $\omega_y$ also act on $F^{(2)}$ in a natural way. Finally, we come to the quotient surface $S := F^{(2)}/\langle \omega_x, \omega_y \rangle$.
Is there a (simple) way to determine if $S$ is a rational surface or not? It is reasonale to apply Castelnuolo's rationality criterion, but I don't know how to compute the irregularity $q$ and second plurigenus $P_2$ of $S$. Is it necessary to construct explicit equation(s) of $S$ to answer my question?
I am also interested in a similar simpler question when $F\!: x^4 + y^4 + 1 = 0$ is the Fermat quartic. Instead of $\omega_x$, $\omega_y$ we have the two involutions $i_x(x,y) := (-x,y)$ and $i_y(x,y) := (x,-y)$. As above, the quotient $F/\langle i_x, i_y \rangle$ is the diagonal conic. What about (non-)rationality of the quotient surface $S := F^{(2)}/\langle i_x, i_y \rangle$?
Thank you in advance for your help!