The answer is $k = \frac{16}{9}$.
We start with a lemma.

Lemma 1. If real numbers $s$ and $t$ are not all positive, then
$$\frac{4}{3}(s^2 - s + 1)(t^2 - t + 1) \geq (st)^2 - st + 1. \quad (*)$$
Proof: Without loss of generality, we assume that $s \geq t$.
We first assume that $s \geq 0 \geq t$. Setting $u = -t$, $(*)$ reads
$$\frac{4}{3}(s^2 - s + 1)(u^2 + u + 1) \geq (su)^2 + su + 1,$$
or
$$4(s^2 - s + 1)(u^2 + u + 1) \geq 3s^2u^2 + 3su + 3.$$
Expanding the left-hand side gives
$$4s^2u^2 + 4s^2u - 4su^2 - 4su + 4s^2 + 4u^2 - 4s + 4u + 4 \geq 3s^2u^2 + 3su + 3,$$
or
$$s^2u^2 + 4u^2 + 4s^2 + 1 + 4s^2u + 4u \geq 4su^2 + 4s + 7su$$
which is evident as $s^2u^2 + 4u^2 \geq 4su^2$, $4s^2 + 1 \geq 4s$, and $4s^2u + 4u \geq 8su \geq 7su$.

We second assume that $0 \geq s \geq t$. Let $v = -s$. By our previous argument, we have
$$\frac{4}{3}(v^2 - v + 1)(t^2 - t + 1) \geq (vt)^2 - vt + 1.$$
It is clear that $t^2 - t + 1 > 0$, $s^2 - s + 1 \geq v^2 - v + 1$, and $(vt)^2 - vt + 1 \geq (st)^2 - st + 1$. Combining the last four inequalities gives $(*)$, and this completes the proof of the lemma.

Now we show that if $x, y, z$ are not all positive real numbers, then
$$\frac{16}{9}(x^2 - x + 1)(y^2 - y + 1)(z^2 - z + 1) \geq (xyz)^2 - xyz + 1. \quad (**)$$
We consider three cases.

a. We assume that $y \geq 0$. Setting $(s, t) = (y, z)$ and then $(s, t) = (x, yz)$ in the lemma gives the desired result.

b. We assume that $0 \geq y$. Setting $(s, t) = (x, y)$ and then $(s, t) = (xy, z)$ in the lemma gives the desired result.

Finally, we confirm that the minimum value of $k$ is $\frac{16}{9}$ by noting that the equality holds in $(**)$ when $(x, y, z) = (\frac{1}{2}, \frac{1}{2}, 0)$.

We establish $(**)$ by showing
$$g(z) = \frac{16}{9}(x^2 - x + 1)(y^2 - y + 1)(z^2 - z + 1) - (xyz)^2 + xyz - 1 \geq 0.$$
Note that $g(z)$ is a quadratic in $z$ whose axis of symmetry (found by comparing the linear and quadratic terms) is at
$$\begin{aligned} z &= \frac{1}{2} - \frac{9}{32} \cdot \frac{xy}{(x^2 - x + 1)(y^2 - y + 1)} \\ &= \frac{1}{2} - \frac{9}{32} \cdot \frac{1}{\left(x + \frac{1}{x} - 1\right)\left(y + \frac{1}{y} - 1\right)}. \end{aligned}$$
For any $t$, we have $|x + \frac{1}{x} - 1| \ge 1$, so the absolute value of the second quantity on the right-hand side of the above equation is at most $\frac{9}{32}$, which is less than $\frac{1}{2}$. That is, the axis of symmetry occurs to the right side of the $y$-axis, so we only decrease the difference between the sides by replacing $z$ by $0$. But when $z = 0$, we only need to show
$$g(0) = \frac{16}{9}(x^2 - x + 1)(y^2 - y + 1) - 1 \geq 0,$$
which is evident as $t^2 - t + 1 = (t - \frac{1}{2})^2 + \frac{3}{4} \ge \frac{3}{4}$.

This is the Calculus version of the second solution. We maintain the same notation as in the second solution. We have
$$\frac{dg}{dz} = \frac{16}{9}(2z - 1)(x^2 - x + 1)(y^2 - y + 1) - 2zx^2y^2 + xy$$
or
$$\frac{dg}{dz} = 2z \left[ \frac{4}{3}(x^2 - x + 1)\frac{4}{3}(y^2 - y + 1) - x^2y^2 \right] + \left[ xy - \frac{4}{3}(x^2 - x + 1)\frac{4}{3}(y^2 - y + 1) \right]. \quad (\dagger)$$
It is evident that
$$\frac{4}{3}(t^2 - t + 1) \geq t^2 \geq 0$$
as it is equivalent to $t^2 - 4t + 4 = (t - 2)^2 \ge 0$. It follows that
$$2z \left[ \frac{4}{3}(x^2 - x + 1)\frac{4}{3}(y^2 - y + 1) - x^2y^2 \right] \le 0;$$
that is, the first summand on the right-hand side of $(\dagger)$ is not positive. It is also evident that
$$\frac{4}{3}(t^2 - t + 1) \geq t$$
as it is equivalent to $4t^2 - 7t + 4 = 4\left(t - \frac{7}{8}\right)^2 + \frac{15}{16} > 0$. If $y \ge 0$, then multiplying the inequalities
$$\frac{4}{3}(x^2 - x + 1) \ge x \ge 0 \quad \text{and} \quad \frac{4}{3}(y^2 - y + 1) \ge y \ge 0$$
gives
$$\frac{4}{3}(x^2 - x + 1)\frac{4}{3}(y^2 - y + 1) - xy \ge 0.$$
If $y < 0$, then $xy < 0$, and so
$$\frac{4}{3}(x^2 - x + 1)\frac{4}{3}(y^2 - y + 1) \ge 0 \ge xy.$$
In either case, we have shown that the second summand in $(\dagger)$ is also negative. We conclude that $\frac{dg}{dz} \le 0$ for $z \le 0$. Hence $g(z)$ reaches minimum when $z = 0$, and we can finish as we did in the second solution.