Here is a solution with a boring PoP calculation. Let $(DFQ) \cap AB=R, (DEQ) \cap AC=S$. We have that $BR.BF=BD.BX$ and $CS.CE=CY.CD$, so it is easy to calculate that we have to prove $\frac {BR} {CS}= \frac {AB} {AC}$. This is easy to see as $AF. AR=AQ. AD=AE.AS$ gives that $\frac {AR} {AS}= \frac {AB} {AC}$, done.

Let $H$ be the orthocenter of $ABC$ .We have $(QDF),(HDF),(DFA)$ are coaxal then denote the power of point P with respect $(DAF)$ as $ p_1(P)$ ; the power of point P with respect $(DHF)$ as $ p_1(P) $ $\frac{ p_1(Q)}{ p_2(Q)}=\frac{ p_1(X)}{ p_2(X)}$ implies $ \frac{QA }{QH }= \frac{XC }{XB }$; similarly we get $ \frac{QA }{QH }= \frac{YB }{YC }$ so $ \frac{XC }{XB }= \frac{YB }{YC }$ consider $Y'$ the reflection of $Y$ in the midpoint of $CB$ then $ \frac{XC }{XB }= \frac{Y'C }{Y'B }$ but $X,Y'$ are in the same side wrt $B$ and $C$ then $Y'=X$ because the apollonian circle cut the line $BC$ at two points in different side wrt one of the point.

Nice Problem WLOG $B,Y,X,C$ lies in this order. Let $(QDF)\cap AB=M\neq F,(QDE)\cap AC=N\neq E$, and let $H$ be the orthocenter of triangle $ABC$. Now, we bash power. $$BD\cdot BX=BF\cdot BM,BD\cdot BC=BF\cdot BA\implies BX=\frac{BC}{BA}\cdot BM$$$$CD\cdot CY=CE\cdot CN,CD\cdot CB=CE\cdot CA\implies CY=\frac{CB}{CA}\cdot CN$$So, it's equivalent to prove that $\frac{BM}{BA}=\frac{CN}{CA}$ or $MN\parallel BC$. Note that $$\measuredangle QMA=\measuredangle QMF=\measuredangle QDF=\measuredangle HDF=\measuredangle HBF=\measuredangle EBA.$$Thus, $MQ\parallel BE$ so $MQ\perp AN$. Similarly, $NQ\parallel CF$ so $NQ\perp AM$. Hence, $Q$ is the orthocenter of $AMN$ which mean that $AQ\perp MN$ but $AQ\perp BC$ too, so $MN\parallel BC$, as desired.

Let $U$ be the second intersection of $(QDF)$ and $AB$, $V$ is the second intersection of $(QDE)$ and $AC$. Note that $\angle BXU=\angle BFD=\angle ACB$, so $UX \parallel AC$. Similarity $VY\parallel AB$. Moreover $$\overline {AF}\cdot \overline {AU}=\overline {AQ}\cdot \overline {AD}=\overline {AE}\cdot \overline {AV}.$$That implies $$\dfrac{\overline {AU}}{\overline {AV}}=\dfrac{\overline {AE}}{\overline {AF}}=\dfrac{\overline {AB}}{\overline {AC}}.$$Then $UV\parallel BC$, so $UVYB$ is a parallelogram, lead to $BU=VY$. Thus $\triangle BUX \sim \triangle YVC$, so $BX=CY$.

By simple angle chasing $FHQ \sim FBX$ and $EHQ \sim ECY$. Thus we have $\frac{BX}{HQ}=\frac{BF}{FH}=\frac{CE}{EH}=\frac{YC}{HQ}$.

Denote H the orthocenter of ABC Since $\angle QEY = \angle QFX = 90$ so we obtain $\angle BFX = \angle CFQ , \angle CEY = \angle QEB$ so by Sin rule in CEY, BFX we have : CY= sin(CEY)/ sin(C ). EY , BX= sin(BFX)/sin(B). FX so it suffices to prove that Sin(B)/sin(C ) . Sin(CEY) /sin(BFX) = FX/EY From Sin rule in QHE , QHF we obtain sin(B)/sin(C ) . Sin(CEY)/sin(BFX) = QF/QE So it is enough to prove that QF/QE = FX/EY or equivalently tan(FDQ) = tan(EDQ) Which is clear.