An equation $f(x,y,z) = 0$ is said to be a cyclic identity if $$\left( \frac{\partial x}{\partial y} \right) \left( \frac{\partial y}{\partial z} \right) \left( \frac{\partial z}{\partial x} \right) = -1.$$
I have been trying to figure out how to prove $f(x,y,z)=x^{2}+y-z = 0$ is a cyclic identity and even wrote down all the possible derivatives of the equations $z=x^{2}+y$ and $x=\sqrt{z-y}$ and $y=z-x^{2}$
However, I can not find anything to cancel out the x's and square root values. Can someone please help me work through this, I don't know what else I can do.
I follow this definition of cyclic identity:
https://en.wikipedia.org/wiki/Triple_product_rule
Write $f(x,y,z) = x^2+y-z = 0$. Note for example that $$x = \pm \sqrt{z-y}.$$ Compute $$\frac{\partial x}{\partial y} = \frac{ - \text{sign}(x)}{2 \sqrt{z-y}}$$ and $$\frac{\partial y}{\partial z} = 1, \, \frac{\partial z}{\partial x} = 2x.$$ Now multiply all partial derivatives, and find $$\frac{\partial x}{\partial y} \frac{\partial y}{\partial z} \frac{\partial z}{\partial x} = -\text{sign}(x) \frac{x}{\sqrt{z-y}} = -\text{sign}(x) \frac{x}{\sqrt{x^2}} = -\text{sign}(x) \frac{x}{|x|} = -\text{sign}(x) \text{sign}(x) = -1.$$