Solution:
Answer: min $\sqrt{3}$ when all equal.

Let us consider $z$ to be fixed and focus on $x$ and $y$. Put $f(x, y, z) = \dfrac{xy}{z} + \dfrac{yz}{x} + \dfrac{zx}{y}$. We have $f(x, y, z) = \dfrac{p}{z} + \dfrac{z(1 - z^2)}{p} = \dfrac{p + k^2 / p}{z}$, where $p = xy$, and $k = z\sqrt{1 - z^2}$. Now $p$ can take any value in the range $0 < p \leq (1 - z^2)/2$. The upper limit is achieved when $x = y$.

We have $p + k^2 / p = \dfrac{(p - k)^2}{p}$. For $p \leq k$, $(p - k)$ and $1/p$ are both decreasing functions of $p$, so $p + k^2 / p$ is a decreasing function of $p$. Thus if $p$ is restricted to the interval $(0, h]$, then for $k \leq h$ the minimum value of $p + k^2 / p$ is $2k$ and occurs at $p = k$. For $k \geq h$ the minimum is $h + k^2 / h$ and occurs at $p = h$.

We have $h = (1 - z^2)/2$, $k = z\sqrt{1 - z^2}$. So $k \leq h$ iff $z \leq 1/\sqrt{5}$. So if $z \leq 1/\sqrt{5}$, then $f(x, y, z) \geq 2k / z = 2\sqrt{1 - z^2} \geq 2\sqrt{1 - 1/5} = 4/\sqrt{5} > \sqrt{3}$.

If $z > 1/\sqrt{5}$, then the minimum of $f(x, y, z)$ occurs at $x = y$ and is $x^2 / z + z + z = (1 - z^2)/(2z) + 2z = 3z/2 + 1/(2z) = (\sqrt{3}/2)(z\sqrt{3} + 1/(z\sqrt{3})) \geq \sqrt{3}$ with equality at $z = 1/\sqrt{3}$ (and hence $x = y = 1/\sqrt{3}$ also).