Let $f : \mathbb{R}^n \rightarrow \mathbb{R}^n$. Then
$$\frac{\partial f^i}{\partial x^j} = (\nabla f)^i_j$$
where $\nabla f$ is the Jacobian matrix of $f$. When reading this paper I came across the expression
$$\frac{\partial x^i}{\partial f^j}$$
Should I interpret this as
$$\frac{\partial x^i}{\partial f^j} = ((\nabla f)^{-1})^i_j$$
Imagine you can invert the problem $x^i = x^i(f)$. Clearly
$$ x^i = x^i(f^1(x),\cdots,f^n(x)) $$
Now apply the chain rule
$$ \frac{\partial x^i}{\partial x^j} = \frac{\partial x^i}{\partial f^k} \frac{\partial f^k}{\partial x^j} = (\nabla_f x)^{i}_{\;k}(\nabla_x f)^{k}_{\;j} = \delta^i_j $$
That means that
$$ \mathbb{1} = (\nabla_f x) (\nabla_x f) $$
Or in other words
$$ \nabla_f x = (\nabla_x f)^{-1} $$