Also my problem

We have $\angle DBX = 180^{\circ} - \angle BYX = \angle CYX = \angle DAX$, so $DABX$ is a cyclic quadrilateral. Therefore, $\overline{ZA} \cdot \overline{ZD} = \overline{ZB} \cdot \overline{ZX}$, so $Z$ lies on the radical axis of $(AED)$ and $(BEC)$. It is well-known that $(AED)$ and $(BEC)$ are tangent to each other, so $Z$ must lie on the internal common tangent of those two circles, hence $ZE$ is tangent to $\omega$.

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Proof of $BE$ is tangent to the circumcircle of $\triangle BXY \implies EZ$ is tangent to $\omega$. $$\angle XEC = \angle XBC = \angle XZA \implies \ \text{XZAE is cyclic.}$$\begin{align*} \angle XEZ & = \angle XAZ \\ & = \angle XYB \\& = \angle XBE \end{align*}Hence $ZE$ is tangent to $(BEX) \equiv \omega$. Proof of $EZ$ is tangent to $\omega \implies BE$ is tangent to the circumcircle of $\triangle BXY$. $$\angle XEC = \angle XBC = \angle XZA \implies \ \text{XZAE is cyclic.}$$\begin{align*} \angle XBE & = \angle XEZ \\ &= \angle XAZ \\ &= \angle XYB \end{align*}Hence $BE$ is tangent to $(BXY)$.