Define $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ by $f(x,y) = (x-y, xy)$. I have to find the points where $f$ is not locally invertible. The Jacobian is nonsingular at all points NOT of the form (x, -x), so by the inverse function theorem, $f$ is locally invertible at those points. Now, I know that at each point of the form $(x, -x)$ the function is not locally invertible, because every neighborhood of $(x,-x)$ contains points of the form $(x + t,-x + t)$ and $(x - t, -x - t)$ for some $t > 0$. But $f$ maps both of those points to the same point.
But one thing that is bothering me is that I thought I found an inverse for $f$. I set $f(x,y) = (x-y, xy) = (u,v)$. So $x -y =u$ and $xy = v$. The first equation implies that $x = u + y$. Substituting this into the second equation gives $y^2 + yu - v = 0$. It's easy to check that the discriminant of this is non-negative, and thus it always have a solution. Solving gives $y = \frac{-u \pm \sqrt{u^2 + 4v}}{2}$, which means that $x = \frac{u \pm \sqrt{u^2 + 4v}}{2}$. If I let $g(u,v) = (\frac{u \pm \sqrt{u^2 + 4v}}{2}, \frac{-u \pm \sqrt{u^2 + 4v}}{2})$, then $(f \circ g)(u,v) = (u,v)$. So doesn't this mean that $g = f^{-1}$? But then it looks like I solved for a global inverse, which I just showed above it impossible. Where am I going wrong?