Solved with Erkosfobiladol. Let $(XOMY) \cap BC=P,M$. We have $\angle PXO=90=\angle PYO \implies PX,PY$ are tangent to $(ABC)$. So $PX=PY$. Let $K$ be the altitude from $D$ to $EF$.$K$ is Sharky-Devil Point which means $X,K,I$ are collinear. $sin \angle \frac{C}{2}.DF=FK$ and $sin \angle \frac{B}{2}.DE=KE \implies \frac{FK}{KE}=\frac{FD}{DE}.\frac{sin \angle \frac{C}{2}}{sin \angle \frac{B}{2}}=\frac{IB.sin \angle B}{IC.\sin \angle C}.\frac{sin \angle \frac{C}{2}}{sin \angle \frac{B}{2}}=\frac{IB}{IC}.\frac{cos \angle \frac{B}{2}}{cos \angle \frac{C}{2}}=\frac{BD}{DC}$ Since $X$ is the center of spiral homothethy which sends $[EF]$ to $[CB]$,$\boxed{XFKE \sim XBDC}$ Let $EF \cap BC=T$.By similarity we get $\angle XKT=\angle XKF=\angle XDB=\angle XDT$, so $X,K,D,T$ are cyclic. Let $R$ be the midpoint of $[TD]$.$\angle DKT=90$ which means $R$ is the center of $(XKDT)$. \[(T,D;B,C)=-1\]so $RB.RC=RD^2=RX^2$ so $RX$ is tangent to $(XABC)$. Also we know $PX$ is tangent to $(XABC)$ and $P,R \in BC$. $\implies R=P$ $\implies PY=PX=PD$ $\implies P$ is the center of $(XYD)$ as desired.

Let $L$ be the circumcenter of $\triangle DXY$,$AI\cap \odot(ABC)=W$,$YD\cap \odot(ABC)=V$ $\mathrm{Lemma.1}$ $O,M,W$ are collinear Note that $\angle \dfrac{XB}{XC}=\dfrac{BF}{CE}=\dfrac{BD}{DC}$ Hence $\angle BXW=\angle CXW=\dfrac{1}{2}\angle BAC$ So $BW=BC$ Which mean that $O,M,W$ are collinear $\mathrm{Lemma.2}$ $V,O,M,W$ are collinear Note that $\angle YXM=\angle YOW=2\angle YVW=2\angle YXW$ So $\angle DXM=\angle DYW$ Suppose $XM\cap VW=M'$ Note that $X,D,M',V$ are cyclic Hence $\angle DM"V=90^{\circ}$ Which means that $M'=M$ Hence $V,O,M,W$ are collinear $\mathrm{Finally}$ Note that $\angle XYD=2\angle XYV=2\angle XWV$ So $\angle XDL=90^{\circ}-\dfrac{1}{2}\angle XYD$ Which means that $L,B,C$ are collinear

Sketch of my solution: Let $N= AI \cap (ABC) \ \ ; \ \ S=ON \cap (ABC)$ By Sharky-Devil lemma, we obtain $X,D,N$ are collinear. Then, let $Y'=SD \cap (ABC)$. By angle-chasing, $Y' \equiv Y$ $L=(XOMY) \cap BC$, again angle-chasing implies that $L$ is circumcenter of $DXY.$