Clearly, all summands have the same sign, so they all must be positive. So either all of $x,y,z$ are positive or exactly two of them negative. But in the latter case we can replace these by their absolute value and get another solution. So it suffices to find all solutions with $x \ge y \ge z >0$. But $\frac{xy}{z} <3$ and hence $xy<3z$ so $x<3$. So $x=2$ or $x=1$. If $x=1$, then $x=y=z=1$ which is indeed a solution. If $x=2$, we have three options $y=z=2, y=2, z=1$ and $y=z=1$. But none of them satisfies the equation. Hence we have (again including the signs) four solutions: $(1,1,1)$ and $(1,-1,-1)$ and $(-1,1,-1)$ and $(-1,-1,1)$.

Use A.M >=G.M. to this expression. we get x*y/z + y*z/x + z*x/y >= 3*sqrt (x*y*z) hence, 3*x*y*z = 3, Implies x*y*z = 1 so, the solutions are (1,1,1) , (1,-1,-1) ,(-1,-1,1) and (-1,1,-1).

debasish123 wrote: Use A.M >=G.M. to this expression. we get x*y/z + y*z/x + z*x/y >= 3*sqrt (x*y*z) hence, 3*x*y*z = 3, Implies x*y*z = 1 so, the solutions are (1,1,1) , (1,-1,-1) ,(-1,-1,1) and (-1,1,-1). You cannot use AM - GM for (possibly) negative numbers