If $f: \mathbb{R}^{2} \to \mathbb{R}^{2}$ is given by:
$f(x,y)= \begin{cases} (x,y-x^{2}) & if & x^{2} \leq y \\ (x,\frac{y^{2}-x^{2}}{x^{2}}) & if & 0 \leq y \leq x^{2} \\ -f(-x,-y) & if & y \leq 0 \end{cases}$
To prove the Jacobian is the identity it must be proved that $df(0,0)=I_{2 \times 2}$. If $ x^{2} \leq y$, then $f(x,y)=(x,y-x^{2})$. So by definition of Jacobian matrix we got $$df(x,y)=\begin{bmatrix} 1 & 0 \\ -2x & 1 \end{bmatrix}.$$
So it follows that $df(0,0)=I_{2 \times 2}$ as required.
The problems begins considering the case where, $(x,y) \in \mathbb{R}^{2}$ are such $0 \leq y \leq x^{2}$, then $f(x,y)=(x,\frac{y^{2}-x^{2}}{x^{2}})$. Obtaining the following Jacobian matrix:
$$df(x,y)=\begin{bmatrix} 1 & 0 \\ \frac{-2y^{2}}{x^3} & \frac{2y}{x^2} \end{bmatrix}.$$
But $df(0,0)$ doesnt make sense since $\frac{-2y^{2}}{x^3}$ and $\frac{2y}{x^2}$ doesn make sense evaluated at $(0,0)$.
If $(x,y) \in \mathbb{R}^{2}$ are such $y \leq 0 $, then $f(x,y)=-f(-x,-y)$. And as $-y \geq 0$ we can consider the two cases before. If first case hold $x^{2} \leq y$ we are done and if $0 \leq y \leq x^{2}$ the second case hold. But second case troubles me for what I mentioned before.
Any help in order to finish this proof will be aprecciated. Thanks!