This should be a simple question, but I've no idea about a strict proof.
Consider a $2\times2$ map $(x,y)\mapsto(f(x,y),g(x,y))$, where $f:\mathbb{R}^2\to\mathbb{R}$ and $g:\mathbb{R}^2\to\mathbb{R}$ are both continuous (sufficiently smooth, if needed) functions. If the Jacobian matrix of this map is invertible, then the map is locally invertible, which is a result of the inverse function theorem. However, there may be some looser conditions for such an invertibility.
Suppose that in a neighborhood of $(x^*,y^*)=(0,0)$, the function $f>0$ to one side of the nullcline of $f$ and $f<0$ to the other side, and the function $g>0$ to one side of the nullcline of $g$ and $g<0$ to the other; moreover, the two nullclines of $f$ and $g$ intersect at $(0,0)$, transversally.
Does it guarantee that the map $(x,y)\mapsto(f,g)$ is an invertible one?
I believe this is true, but still wonders how to prove it.
Let $h(x) = \begin{cases} x^2 & x \ge 0 \\ -x^2 & x < 0 \end{cases}$.
Let $q(x) = \begin{cases} \sin \frac{1}{x} & x \ne 0 \\ 0 & x = 0 \end{cases}$.
Then I believe that $$ k(x) = (xq(x) + x) + h(x) $$ is strictly positive to the right of $0$, and strictly negative to the left of $0$, but is not invertible in any neighborhood of $0$.
Letting $f(x, y) = k(x); g(x, y) = y$, the map $(x, y) \to (f(x, y), g(x, y))$ is not invertible in any neighborhood of $(0,0)$, despite satisfying your conditions.