Consider the mapping $f: \mathbb R^2 \backslash \{(0,0)\} \to \mathbb R^2$ given by $$\begin{aligned} f(x,y) = \begin{pmatrix} (x^2-y^2)/(x^2+y^2) \\ xy/(x^2+y^2) \end{pmatrix} \end{aligned}$$ Does $f$ have a local inverse at every point of $\mathbb R^2$?
Update: The interesting fact with this function is that we can not use the inverse function theorem to decide whether it is locally injective, since for all points $(x, y)$ we have $Jf (x, y) = 0$. Indeed,
\begin{align} \det Jf(x,y) = & \det \begin{pmatrix} \frac{\partial}{\partial x} \frac{ (x^2-y^2)}{(x^2+y^2)} & \frac{\partial}{\partial y} \frac{ (x^2-y^2)}{(x^2+y^2)} \\ \frac{\partial}{\partial x} \frac{ xy}{(x^2+y^2)} & \frac{\partial}{\partial y} \frac{xy}{(x^2+y^2)} \end{pmatrix} \\ =& \det \begin{pmatrix} \frac{ 2x(x^2+y^2)-(x^2-y^2)2x}{(x^2+y^2)^2} & \frac{-2y(x^2+y^2)-(x^2-y^2)2y}{(x^2+y^2)^2} \\ \frac{y(x^2+y^2)-xy(2x) }{(x^2+y^2)^2} & \frac{ x(x^2+y^2)-xy(2y) }{(x^2+y^2)^2} \end{pmatrix} \\ =& \det \begin{pmatrix} \frac{ 4xy^2}{(x^2+y^2)^2} & \frac{-4x^2y}{(x^2+y^2)^2} \\ \frac{-x^2y+y^3 }{(x^2+y^2)^2} & \frac{ x^3-xy^2 }{(x^2+y^2)^2} \end{pmatrix} \\ =& \frac{ 4xy}{(x^2+y^2)^2} \det \begin{pmatrix} y & -x \\ \frac{-x^2y+y^3 }{(x^2+y^2)^2} & \frac{ x^3-xy^2 }{(x^2+y^2)^2} \end{pmatrix} \\ =& \frac{ 4xy}{(x^2+y^2)^2} \cdot \frac{ 1}{(x^2+y^2)^2} \det \begin{pmatrix} y & -x \\ {-x^2y+y^3 } & { x^3-xy^2 } \end{pmatrix} \\ =& \frac{ 4xy}{(x^2+y^2)^2} \cdot \frac{ 1}{(x^2+y^2)^2} [ y({ x^3-xy^2 }) +x({-x^2y+y^3 })] \\ =& \frac{ 4xy}{(x^2+y^2)^2} \cdot \frac{ 1}{(x^2+y^2)^2} [ x^3y-xy^3-x^3y+xy^3] \\ =& \frac{ 4xy}{(x^2+y^2)^2} \cdot \frac{ 1}{(x^2+y^2)^2} \cdot 0 \\ =&0 \end{align}
This map is not locally injective for all points in your domain. For $x = y=t$ we have \begin{align} f(t,t)=& \begin{pmatrix} 0/(t^2+t^2) \\ t^2/(t^2+t^2) \end{pmatrix} = \begin{pmatrix} 0 \\ \frac{1}{2} \end{pmatrix} \end{align} Therefore, this function is not locally invertible in any neighborhood of points $ x = y $.