So I have seen in the derivation of Euler chain Equation (i.e. $\displaystyle \frac{\partial x}{\partial y} \cdotp \frac{\partial y}{\partial z} \cdotp \frac{\partial z}{\partial x} \ =\ -1$ )
where we use $\displaystyle \frac{\partial y}{\partial x} \ =\ \frac{1}{\partial x/\partial y}$
So, I have a contradiction.
Let's say $x = rcos(\theta)$ -------- (1)
and $y = rsin(\theta)$, so we have $x^2+y^2=r^2$ ---------(2)
$\displaystyle \frac{\partial r}{\partial x} \ = cos(\theta)$ from eqn(2)
And $\displaystyle \frac{\partial x}{\partial r} \ = cos(\theta)$ from eqn(1).
Where is the problem?
If $\displaystyle \frac{\partial y}{\partial x} \ =\ \frac{1}{\partial x/\partial y}$ is false, then what are the conditions where it's true?
Here, you have functions of two variables, and when there is ambiguity, you have to indicate which variable is fixed when you compute a partial derivative.
For example, you can fix $\theta$ when you derive $x=r\cos\theta$ and get $\displaystyle \Big( \frac{\partial r}{\partial x} \Big)_\theta\ = \frac{1}{cos(\theta)}.$
Or you can fix $y$ when you derive $x^2+y^2=r^2$ and get $\displaystyle 2x \Big(\frac{\partial x}{\partial r} \Big)_y = 2r$, so $\displaystyle \Big(\frac{\partial x}{\partial r} \Big)_y = \frac{r}{x} = \frac{1}{cos(\theta)}.$