= Solution
Work locally where the needed derivatives of the <level set> can be represented by the <implicit function theorem>. Differentiating while holding the indicated coordinate fixed gives
$$
\left(\frac{\partial x}{\partial y}\right)_z=-\frac{f_y}{f_x},\qquad
\left(\frac{\partial y}{\partial z}\right)_x=-\frac{f_z}{f_y},\qquad
\left(\frac{\partial z}{\partial x}\right)_y=-\frac{f_x}{f_z}.
$$
Multiplying the three <implicit differentiation> formulas proves
$$
\boxed{\left(\frac{\partial x}{\partial y}\right)_z
\left(\frac{\partial y}{\partial z}\right)_x
\left(\frac{\partial z}{\partial x}\right)_y=-1.}
$$
This <cyclic partial derivative identity> is asserted where all three finite coordinate derivatives exist, in particular when $f_xf_yf_z\ne0$. It is not a globally defined product at points where one of the necessary coordinate parametrizations is singular.
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