If x and y are acute angles such that

Given $\cos x=\frac{13}{14}$ and $\cos y=\frac{1}{7}$

Now we will calculate value of sinx and siny

$\sin x=\sqrt{\left(1-\cos ^{2} x\right)} \Rightarrow \sqrt{\left(1-\left(\frac{13}{14}\right)^{2}\right)}=\sqrt{\left(\frac{196-169}{196}\right)} \Rightarrow \sqrt{\left(\frac{27}{196}\right)}=\frac{3 \sqrt{3}}{14}$

siny $=\sqrt{\left(1-\cos ^{2} y\right)} \Rightarrow \sqrt{\left(1-\left(\frac{1}{7}\right)^{2}\right)}=\sqrt{\left(\frac{49-1}{49}\right)} \Rightarrow \sqrt{\left(\frac{48}{49}\right)}=\frac{4 \sqrt{3}}{7}$

Hence,

$\cos (x-y)=\cos x \cdot \cos y+\sin x \cdot \sin y$

$=\frac{13}{14} \cdot \frac{1}{7}+\frac{3 \sqrt{3}}{14} \cdot \frac{4 \sqrt{3}}{7} \Rightarrow \frac{13+36}{98}=\frac{49}{98}$

$\cos (x-y)=\frac{1}{2}$

$x-y=\frac{\pi}{3}$