Let $xyz = a$. Then
$$\begin{aligned} (xy - z^2)(yz - x^2)(zx - y^2) &= \\ &= \left(\frac{a}{z} - z^2\right) \left(\frac{a}{x} - x^2\right) \left(\frac{a}{y} - y^2\right) = \frac{1}{a}(a - z^3)(a - x^3)(a - y^3) = \\ &= \frac{1}{a}(a^3 - (x^3 + y^3 + z^3)a^2 + (x^3y^3 + y^3z^3 + z^3x^3)a - x^3y^3z^3) = \\ &= [x^3y^3z^3 = a^3] = (x^3y^3 + y^3z^3 + z^3x^3) - (x^3 + y^3 + z^3)a = A - Ba, \quad (1) \end{aligned}$$
where
$$A = x^3y^3 + y^3z^3 + z^3x^3 \quad \text{and} \quad B = x^3 + y^3 + z^3.$$
Let $xy^2 + x^2y + yz^2 + y^2z + zx^2 + xz^2 = C$. Find the values of $A$, $B$, $C$. First,
$$x^2 + y^2 + z^2 = (x + y + z)^2 - 2(xy + yz + zx) = 1 - 2 \cdot (-1) = 3.$$
Secondly,
$$B = x^3 + y^3 + z^3 = (x+y+z)(x^2+y^2+z^2) - C = (-1) \cdot 3 - C = -3 - C. \quad (2)$$
Further,
$$C = (x + y + z)(xy + yz + zx) - 3xyz = -1 \cdot (-1) - 3a = 1 - 3a.$$
From (2) it follows that
$$B = -3 - (1 - 3a) = 3a - 4.$$
Then
$$x^2y^2 + y^2z^2 + z^2x^2 = (xy + yz + zx)^2 - 2xyz(x+y+z) = (-1)^2 - 2 \cdot (-1) = 2a + 1.$$
So,
$$A = (xy+yz+zx)(x^2y^2+y^2z^2+z^2x^2)-Cxyz = (-1)(2a+1)-(1-3a)a = 3a^2-3a-1.$$
Substituting $A$ and $B$ in (1), we obtain
$$A - Ba = 3a^2 - 3a - 1 - (3a - 4)a = a - 1 = xyz - 1,$$
as required.