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Let $a=x,$ $b=y,$ and $c=-z.$ Then $x=a,$ $y=b,$ and $z=-c,$ so the given equations become $$\begin{array}{rl}a+b+c& =0,\\ -ac-ab-bc& =27,\\ -abc& =54.\end{array}$$We re-write these as $$\begin{array}{rl}a+b+c& =0,\\ ab+ac+bc& =-27,\\ abc& =-54.\end{array}$$Then by Vieta's formulas, $a,$ $b,$ and $c$ are the roots of $$t^3-27t+54=0,$$which factors as $(t-3{)}^2(t+6)=0.$ Therefore, $a,$ $b,$ $c$ are equal to 3, 3, and $-6,$ in some order.

This leads to the solutions $(x,y,z)=(3,3,6),$ $(3,-6,-3),$ $(-6,3,-3),$ for a total of $\boxed{3}$ solutions.