I got two critical points:
$ \left(-1,\dfrac{2}{5},0\right)$ and $\left (\dfrac{25}{8}, \dfrac{5}{18},-\dfrac{5}{8}\right)$.
This is how I got these. First set each of the right hand sides of your three equation to zero as Chisigma said noting that $y \ne 0$ due to the division by $y$. From the second we have
$-\dfrac{1}{4}\left(x+4z-4 \right) y+\dfrac{1}{4}\,x-\dfrac{1}{4}\,{x}^{2}-\dfrac{5}{4}\,xz+z-{z}^{2} = 0$which we can solve for $y$ provided that $x+4z-4 \ne 0 $ so we consider this case first.
If $x = 4 - 4z$ then the second equation becomes $3z-3 = 0$ giving $z = 1$ and in turn $x = 0$. The first and third equations becomes $-3$ and $5$ respectively which is inadmissible since these are both to be zero. Thus, we can conclude $x+4z-4 \ne 0 $.
Solve the second equation for $y$ gives
$y = \dfrac{(x+4 z) (x+z-1)}{x+4 z-4}$ noting that $x + 4z \ne 0$ as this would give $y = 0$.
Simplifying equations (1) and (3) gives
$-{\dfrac {{x}^{2}+9\,xz+x+12\,{z}^{2}}{x+4\,z}} = 0$ and $\,{\dfrac { \left( x+5\,z \right) z}{x+4\,z}}=0$.
These second of these gives rise to two cases and the critical point fall out.
