I was reading the solution to this problem:
If $x, y, z$ are real numbers and $x+y+z=xyz$ then
$x(1 − y^2 )(1 − z^2 ) + y(1 − z^2 )(1 − x^2 ) + z(1 − x^2 )(1 − y^2 ) = 4xyz$
The solution is to divide both sides by $xyz$. Observing the resemblance to the tan double angle formula, the proof substitutes $x = \tan A, y = \tan B, z = \tan C$ where A, B,C are angles of a triangle. Later on, the proof uses the fact that $A+B+C=\pi$.
But if $A,B, $and $ C$ are angles of the same triangle it follows that the triangle inequality must hold. This puts a restriction on the values of $x, y$, and $z$ that is not specified in the problem statement. The proof, therefore, is not valid for all real numbers $x, y$, and $z$.
Source: Putnam and Beyond pg 239
One can simply prove
$(x(1-y^2)(1-z^2)+y(1-x^2)(1-z^2)+z(1-x^2)(1-y^2)-4xyz) = (yz+xz+xy-1)(xyz-z-y-x)$
thus your long equation is zero with the help of $(xyz-z-y-x)=0$ your condition.