Easy angle/length chasing. Let $AC=b$ and $AB=c$. We have $\angle ABE = 30^{\circ}$ and $\angle ACF = 30^{\circ}$. Thus, we have $AE = \frac{c}{2}$ and $AF=\frac{b}{2}$. Therefore, $CE=b-\frac{c}{2}$ and $BF=c-\frac{b}{2}$. Subtracting these two yields $CE-BF = \frac{3}{2} b - \frac{3}{2} c = \frac{3}{2} (AC-AB)$.

Easy Length Chasing Solution Solution: Notice that, \[AE = \frac{AB}{2} \text{ [Sine Law]}\]\[AF = \frac{AC}{2}\] \[AC - AB = (\frac{AB}{2} + CE) - (\frac{AC}{2} + BF)\]\[CF - BF = AC - AB - \frac{1}{2}AB + \frac{1}{2}AC = \frac{2AC - 2AB - AB + AC }{2} = \frac{3}{2}(AC - AB)\]

CE - BF = (AC-AE) - (AB-AF) = AC - AB/2 - AB + AC/2 = 3/2(AC - AB)

$\color{blue} \boxed{\textbf{SOLUTION}}$ $BE \cap CF \equiv H$ $\color{red} \textbf{Step 1}$ $BF=BHcos30°=\frac{\sqrt 3}{2} BH$ By $\textbf{Similarity}$, $\frac{BF}{BE}=\frac{BH}{AB}$ $\implies AB=\frac{2BE}{\sqrt 3}$ Similarly, $AC=\frac{2CF}{\sqrt 3}$ $\frac{3}{2} (AC-AB) = \sqrt 3 (CF-BE)$ $\color{red} \textbf{Step 2}$ $AE=BEtan30°=\frac{BE}{\sqrt 3}$ $CE=AC-AE= \frac{2CF-BE}{\sqrt 3}$ Similarly, $BF=\frac{2BE-CF}{\sqrt 3}$ So, $CE-BF= \frac{3CF-3BE}{\sqrt 3}= \sqrt 3(CF-BE)$ Which is desired $\blacksquare$