The three variables are functions of t, a parameter, such that x(0)=y(0)=1=-z(0) and x'=dx/dt, etc.
x'=6y, y'=x+z, z'=x+y
y"=x'+z'=6y+x+y=7y+x
y"'=7y'+x'=7y'+6y, y"'-7y'-6y=0, which is a characteristic equation with roots m1, m2, m3 where:
(m+1)(m+2)(m-3)=m^3-7m-6=0, so m1=-1, m2=-2 and m3=3.
From this we see that y=Ae^-t+Be^-2t+Ce^3t. y(0)=1 so A+B+C=1.
[PROOF OF GENERAL SOLUTION:
y'=-Ae^-t-2Be^-2t+3Ce^3t, y"=Ae^-t+4Be^-2t+9Ce^3t,
y'"=-Ae^-t-8Be^-2t+27Ce^3t +
-7y'=7Ae^-t+14Be^-2t-21Ce^3t +
-6y=-6Ae^-t-6Be^-2t-6Ce^3t =
0]
x'=6y=6Ae^-t+6Be^-2t+6Ce^3t; x=-6Ae^-t-3Be^-2t+2Ce^3t.
x(0)=1 so -6A-3B+2C=1.
z=y'-x=-Ae^-t-2Be^-2t+3Ce^3t+6Ae^-t+3Be^-2t-2Ce^3t and z(0)=-1, so
-A-2B+3C+6A+3B-2C=-1, 5A+B+C=-1. Since B+C=1-A, 5A+1-A=-1, A=-1/2.
3-3B+2C=1, -3B+2C=-2, C=-1-B-5A=3/2-B, so -3B+3-2B=-2, B=1.
Finally, C=3/2-B=1/2, C=1/2.
x=3e^-t-3e^-2t+e^3t;
y=-e^-t/2+e^-2t+e^3t/2;
z=-5e^-t/2+e^-2t+e^3t/2.
All original equations and conditions check out correctly.