If $ABCD$ is a square, then $\frac{y}{x} = \frac{1+\sqrt{3}}{2}$.

Now we prove that $\frac{y}{x} \le \frac{1+\sqrt{3}}{2}$.

Denote by $P_1, Q_1, R_1, S_1$ the midpoints of $DA$, $AB$, $BC$, $CD$, and by $E$, $F$, $G$, $H$ the midpoints of $SP$, $PQ$, $QR$, $RS$. Then $P_1Q_1R_1S_1$ is a parallelogram.

Now draw lines $P_1E$, $S_1E$, and denote by $M$, $N$ the midpoints of $DP$, $DS$. Then

$$\begin{aligned} DS_1 &= S_1N = DN = EM, \\ DP_1 &= P_1M = MD = EN, \end{aligned}$$
and
$$\begin{aligned} \angle P_1DS_1 &= 360^\circ - 60^\circ - 60^\circ - \angle PDS \\ &= 240^\circ - (180^\circ - \angle END) = 60^\circ + \angle END \\ &= \angle ENS_1 = \angle EMP_1. \end{aligned}$$
So we have $\triangle DP_1S_1 \cong \triangle MP_1E \cong \triangle NES_1$. Hence, $\triangle EP_1S_1$ is equilateral.

By the same argument, $\triangle GQ_1R_1$ is also equilateral. Now let $U$, $V$ be the midpoints of $P_1S_1$, $Q_1R_1$, respectively. We then obtain
$$\begin{aligned} EG &\le EU + UV + VG = \frac{\sqrt{3}}{2}P_1S_1 + P_1Q_1 + \frac{\sqrt{3}}{2}Q_1R_1 \\ &= P_1Q_1 + \sqrt{3}P_1S_1 = \frac{1}{2}BD + \frac{\sqrt{3}}{2}AC, \end{aligned}$$
and also
$$FH \le \frac{1}{2}AC + \frac{\sqrt{3}}{2}BD.$$

Taking the sum of these two inequalities, we have $y \le \frac{1+\sqrt{3}}{2}x$, i.e.
$$\frac{y}{x} \le \frac{1+\sqrt{3}}{2}.$$