You could think geometrically:

$xyz=1$ is an irreducible hypersurface in $\mathbb A_{\mathbb C}^3$. A maximal chain of primes in $A$ corresponds to a chain of the form

$$\text{surface} \supset \text{curve} \supset \text{point}$$

So, we are looking for a curve on this hypersurface. The easiest way to obtain such a curve is to look at the intersection with another hypersurface, say $x=1$. This gives rise to the curve $yz=1$ in the plane given by $x=1$. This curve corresponds to the prime ideal $(x-1,yz-1)$ and indeed we have $xyz-1=yz(x-1)+(yz-1) \in (x-1,yz-1)$.

Now we are looking for a point on that curve. That is a no-brainer, take $x=y=z=1$, corresponding to the ideal $(x-1,y-1,z-1)$.

Summing up, we have found the chain

$$(xyz-1) \subset (x-1,yz-1) \subset (x-1,y-1,z-1)$$