Given : $AD$ is a median of $\triangle ABC$ (so $BD = DC$), $X$ lies on $AD$ with $AX : XD = 2 : 3$, and $BX$ produced meets $AC$ at $Y$.
To Prove : $BX = 4\,XY$.

Construction : Through $D$, draw $DZ \parallel BY$ meeting $AC$ at $Z$.

Step 1: In $\triangle BCY$, $D$ is the midpoint of $BC$ and $DZ \parallel BY$.
By the converse of the midpoint theorem, $Z$ is the midpoint of $CY$, and
$$DZ = \frac{1}{2}BY \qquad \ldots (i)$$
Step 2: In $\triangle ADZ$, $XY \parallel DZ$ (since $BY \parallel DZ$).
By AA similarity, $$\triangle AXY \sim \triangle ADZ$$
Step 3: Therefore
$$\frac{XY}{DZ} = \frac{AX}{AD}$$
Step 4: Since $AX : XD = 2 : 3$, we get $AD = AX + XD = 2 + 3 = 5$ parts, so $\dfrac{AX}{AD} = \dfrac{2}{5}$.
$$\Rightarrow \frac{XY}{DZ} = \frac{2}{5} \Rightarrow DZ = \frac{5}{2}XY \qquad \ldots (ii)$$
Step 5: From $(i)$ and $(ii)$ :
$$\frac{1}{2}BY = \frac{5}{2}XY \Rightarrow BY = 5\,XY$$
Step 6: But $BY = BX + XY$, therefore
$$BX + XY = 5\,XY \Rightarrow BX = 4\,XY$$
Hence proved.