We add both inequalities and get \[ \frac{1}{6-z} + \frac{1}{z-y} + \frac{1}{y-x} \le x \] By the arithmetic-harmonic mean inequality for the LHS, this yields \[ x \ge \frac{1}{6-z} + \frac{1}{z-y} + \frac{1}{y-x} \ge \frac{9}{(6-z)+(z-y)+(y-x)} = \frac{9}{6-x}.\] Hence $x(6-x)\ge9$ and $(x-3)^2\le0$ and $x=3$. Hence the arithmetic-harmonic mean inequality holds with equality, so that $6-z=z-y=y-3$, which implies $y=4$ and $z=5$. The unique solution is $(x,y,z)=(3,4,5)$.

Set $z=6-a$, $y=z-b$ and $x=y-c$ where $a,b,c>0$. Then the inequalities become $\frac{1}{b}+\frac{1}{c}\le 2$ and $\frac{1}{a}+2\le 6-a-b-c$. The latter becomes \[4\ge \frac{1}{a}+a+b+c\ge 2+b+c,\] so $b+c\le 2$. But by Cauchy-Schwartz (and various others), $(b+c)\left(\frac{1}{b}+\frac{1}{c}\right)\ge 4$. Hence equality must occur here and in $\frac{1}{a}+a\ge 2$. Since $a,b,c>0$, this forces $a=b=c=1$ and hence $(x,y,z)=(3,4,5)$.

We add both inequalities and get \[ \frac{1}{6-z} + \frac{1}{z-y} + \frac{1}{y-x} \le x \] $\Leftrightarrow$ \[ \frac{1}{6-z} +(6-z)+ \frac{1}{z-y}+(z-y) + \frac{1}{y-x}+(y-x) \le 6 \] On the other hand , \[ \frac{1}{6-z} +(6-z)+ \frac{1}{z-y}+(z-y) + \frac{1}{y-x}+(y-x) \ge 6 \] Hence \[ \frac{1}{6-z} +(6-z)+ \frac{1}{z-y}+(z-y) + \frac{1}{y-x}+(y-x) =6 \] so that $6-z=z-y=y-x=1$, which implies $x=3$ ,$y=4$ and $z=5$. The unique solution is $(x,y,z)=(3,4,5)$.