Consider the coordinate transformation under rotation that \begin{align} y' & = y\cos\phi + z\sin\phi \\ z' & = -y\sin\phi + z\cos\phi , \end{align} we can get \begin{equation} \frac{\partial y'}{\partial y} = \cos\phi . \end{equation}
From the equations, we also have \begin{align} y & = y'\cos\phi - z'\sin\phi \\ z & = y'\sin\phi + z'\cos\phi , \end{align} from which we can get \begin{equation} \frac{\partial y}{\partial y'} = \cos\phi . \end{equation} My question is, why \begin{equation} \frac{\partial y'}{\partial y} \neq \frac{1}{\frac{\partial y}{\partial y'}} \end{equation} in this case?
The 2 transformations in your equation may be written as
$y'=f(y,z)$
$y=g(y',z')$
In your first partial differentiation you are treating $z$ as a constant
$$\frac{\partial y'}{\partial y}=\frac{\partial f(y,z)}{\partial y} \bigg | _{z=\text{constant}}$$
In your second partial differentiation you are treating $z'$ as a constant
$$\frac{\partial y}{\partial y'}=\frac{\partial g(y',z')}{\partial y'} \bigg | _{z'=\text{constant}}$$
As you can see the 2 Partial derivatives are in general not reciprocals of each other.
However The reciprocal rule is true for total derivatives (given that $f$ and $g$ are differentiable and their differentials are non-zero)