Denote $\mathbf{x} \in \mathbb{R}^d $, $f(\mathbf{x}) = (f_1(\mathbf{x}), f_2(\mathbf{x}), \cdots, f_d(\mathbf{x}))$, $f_i: \mathbb{R}^d \rightarrow \mathbb{R}$, $i = 1, 2, \cdots, d$. I want to proof the fact that $\frac{\partial f(\mathbf{x})}{\partial \mathbf{x}} \frac{\partial \mathbf{x}}{\partial f(\mathbf{x})} = I$, where $I$ is the identity matrix.
Here is what I have done:
\begin{align*} \left[\frac{\partial \mathbf{x}}{\partial f(\mathbf{x})} \frac{\partial f(\mathbf{x})}{\partial \mathbf{x}} \right]_{ij} &= \sum_{k=1}^d \left[\frac{\partial \mathbf{x}}{\partial f(\mathbf{x})}\right]_{ik} \left[\frac{\partial f(\mathbf{x})}{\partial \mathbf{x}} \right]_{kj} \\ &= \sum_{k=1}^d \frac{\partial x_i}{\partial f_k(\mathbf{x})} \frac{\partial f_k(\mathbf{x})}{\partial x_j} \\ &= \sum_{k=1}^d \frac{\partial x_i}{\partial x_j} \\ &= \left\{ \begin{aligned} 0 & , & \text{if $i \neq j$} \\ d & , & \text{if $i = j$} \end{aligned} \right. \end{align*} which is not equal to $I$.
Where is the problem?
If $x_i = x_i(f_1(\mathbf{x}), f_2(\mathbf{x}),\ldots, f_d(\mathbf{x}))$, then $$\frac{\partial x_i}{\partial x_j} = \sum_{k=1}^d \frac{\partial x_i}{\partial f_k} \cdot \frac{\partial f_k}{\partial x_j}$$ Note that the sum isn't on the left-hand side.