If $\cos(z-x) + \cos(y-z) + \cos(x-y) = -\frac{3}{2}$, then $\sin x + \sin y + \sin z = 0 = \cos x + \cos y + \cos z $.

By the angle-difference identity you mention, the given equation is equivalent to $$\cos x \cos y + \sin x \sin y + \cos y \cos z + \sin y \sin z + \cos z \cos x + \sin z \sin x = -\frac32 \tag{1}$$ Thus, $$3 + 2 (\cos x \cos y + \cdots ) + 2(\sin x \sin y + \cdots ) = 0 \tag{2}$$

But, $$3 = 1 + 1 + 1 = \left(\cos^2 x + \sin^2 x \right) + \left( \cos^2 y + \sin^2 y \right) + \left( \cos^2 z + \sin^2 z\right) \tag{3}$$

So, (2) becomes $$\begin{align} 0 &= \cos^2 x + \cos^2 y + \cos^2 z + 2 \cos x \cos y + 2 \cos y \cos z + 2\cos z \cos x \\ &+ \sin^2 x + \sin^2 y + \sin^2 z + 2 \sin x \sin y + 2 \sin y \sin z + 2 \sin z \sin x \\ &= \left( \cos x + \cos y + \cos z \right)^2 + \left( \sin x + \sin y + \sin z \right)^2 \end{align} \tag{4}$$

Now, the sum of two squares can be zero only if each square is itself zero, and we are done. $\square$

Let $z_1 = \cos x +i \sin x$ etc.

Then $2 \cos (x-y) = \dfrac{z_1}{z_2}+\dfrac{z_2}{z_1}$ etc.

We are given that $\dfrac{z_1}{z_2}+\dfrac{z_2}{z_1}+\dfrac{z_2}{z_3}+\dfrac{z_3}{z_2}+\dfrac{z_3}{z_1}+\dfrac{z_1}{z_3} = -3$

or $\dfrac{z_2+z_3}{z_1}+\dfrac{z_3+z_1}{z_2}+\dfrac{z_1+z_2}{z_3} = -3$

$\Rightarrow \displaystyle \sum_{cyc} \frac{z_2+z_3}{z_1}+1 =0 \Rightarrow (z_1+z_2+z_3)\left(\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}\right) = 0$

Thus $z_1+z_2+z_3 = 0$ or $\displaystyle \frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}=0$

from which the required result follows