First see that if all are equal $x=y=z=w$, then $2x=8x^2$, so $x = 0$ or $x = \frac{1}{4}$. Both solution works. Now suppose that some pair is not equal, wlog $y\neq z$. We see that subtracting $(1) - (2)$ and $(5) - (6)$ we get $$(y-z)(1+y+z+6w) = 0 = (y-z)(1+y+z+6x).$$Because $y\neq z$, then $x = w$. Now we can see that subtracting $(2) - (4)$ and $(3) - (5)$ we get $$(x-y)(1+x+y+6w) = 0 = (x-y)(1+x+y+6z),$$so $x=y$ or $z=w$. Wlog let $x = y$ ($z=w$ case is just cyclic version), then we have $x+z = 8x^2$ and $2x = z^2+x^2 + 6zx = z^2 + 6zx + 9x^2 - 8x^2$, so $(z+3x)^2 = 8x^2+2x = z + 3x$, meaning that $z+3x = 0$ or $z+3x = 1$. So consider two cases: $1.$ $z+3x = 0$, then we have a quadruplet $(t,t,-3t,t)$, putting in both equations we get $-2t=8t^2$, so $t = 0$ (this case has been already considered) and $t = -\frac{1}{4}$. $2.$ $z+3x = 1$, gets us a quadruplet $(t,t,1-3t,t)$, putting again in both equations we get $8t^2+2t-1 = 0$, so $(4t-1)(2t+1)$, meaning that $t=\frac{1}{4}$ (this case also was considered) and $t = -\frac{1}{2}$. So the solutions are $(x,y,z,w) = \{(0,0,0,0); \left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4}\right), \left(-\frac{1}{4},-\frac{1}{4},\frac{3}{4},-\frac{1}{4}\right), \left(-\frac{1}{2},-\frac{1}{2},\frac{5}{2},-\frac{1}{2}\right)\}$ and all cyclic permutation, putting all solutions back to the equations we see that all work.