Given $f : \mathbb{R} \rightarrow \mathbb{R}$, it's possible to show that $f$ is a bijection by considering its derivatives only: if the derivative is always positive or always negative, then the function is strictly monotonic and is therefore invertible.
Is there a similar route to demonstrating that $g: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is invertible given knowledge of the partial derivatives of $g$ only and without assuming that $g$ is linear? I assume that demonstrating that each component $g_i$ is strictly monotonic in each argument $x_j$ would be sufficient - would anything less do? I've considered demonstrating that each component $g_i$ is strictly monotonic in only the corresponding argument $x_i$, but I'm unsure whether this is valid.