How to find that derivative of this function is invertible The following question was part of my real analysis assignment and I having a hard time solving it.

Let $ f : \mathbb{R}^2 \to \mathbb{R}^2 $ be given by the formula $f(x,y)=( 3x+2y+y^2 +|xy| , 2x+3y +x^2+|xy|) $.

I have proved it to be differentiable at $(0,0)$ but I am unable to think whether Df(0,0) is invertible or not?
Jacobian at $(0,0)$ is $5$ so its invertible by inverse function theorem but the answer is it's not invertible.
what I am doing wrong ?
please help.
 A: Let $f_1(x,y)\equiv (f_1(x),f_2(y))$ for each $(x,y)\in\Bbb R^2$. Then the derivative $f’(x,y)$ of the function $f(x,y)$ is a matrix
$\begin{pmatrix}\frac{\partial{f_1}}{\partial{x}} &  \frac{\partial{f_1}}{\partial{y}}\\
\frac{\partial{f_2}}{\partial{x}} &  \frac{\partial{f_2}}{\partial{y}}\end{pmatrix}$,
and $f’(0,0)=\begin{pmatrix} 3 & 2 \\ 2 & 3\end{pmatrix} $ is invertible.
But the conditions of the inverse function theorem does not applicable to $f$ at $(0,0)$, because there are no neighborhood of $(0,0)$ where $f$ is continuously differentiable.
Nevertheless, the function $f$ is invertible on some open neighborhood of $(0,0)$. Let us show this.
Consider auxiliary functions $f_+$ and $f_-$ from $\mathbb{R}^2$ to $\mathbb{R}^2$ such that for each $(x,y)\in\Bbb R^2$ we have
$$f_+(x,y)=(3x+2y+y^2 +xy , 2x+3y +x^2+xy),$$
$$f_-(x,y)=(3x+2y+y^2 -xy , 2x+3y +x^2-xy).$$
Each of these two functions is continuously differentiable and has non-zero Jacobian at $(0,0)$. So, by the inverse function theorem, there exists $0<r<1$ such that these functions are invertible on a closed square $S=\{(x,y)\in\Bbb R^2:|x|+|y|\le r\}$. In particular, the restictions of these functions onto $S$ are injective.
Lemma. The restriction $f_S$ of $f$ onto $S$ is injective.
Therefore,  by the invariance of domain, the restriction of $f_S$ (and so of $f$) onto the interior of $S$ is a homeomorphic embedding, and so invertible.
Proof of Lemma.  The boundary $B$ of $S$ consists of four segments. We claim that $f|B$ is injective. For this suppose $(x,y)\in B$ and consider the following cases.
I)) $x>0$, $y>0$, $y=-x+r$. Then $f(x,y)=(x-xr+r^2+2r , xr-x+3r) $. Thus $f_1(x,y)+f_2(x,y)=r^2+5r$.
II)) $x<0$, $y>0$, $y=x+r$. Then $f(x,y)=(5x+xr+r^2+2r, 5x-xr +3r) $. Thus $f_1(x,y)+f_2(x,y)=10x+r^2+5r$.
III)) $x<0$, $y<0$, $y=-x-r$. Then $f(x,y)= (x+xr+r^2-2r , -x-xr-3r) $. Thus $f_1(x,y)+f_2(x,y)=r^2-5r$.
IV)) $x>0$, $y<0$, $y=x-r$. Then $f(x,y)=(5x-rx+r^2-2r, 5x+xr-3r) $. Thus $f_1(x,y)+f_2(x,y)=10x+r^2-5r$.
Using the above equations for $f_1(x,y)+f_2(x,y)$ and a condition $0<|x|<r$ it is easy to check that $f|B$ is injective.
Also it is easy to check that $f(B)$ is a quadrilateral with vertices $f(0,r)=r(2+r,3)$, $f(r,0)=r(3,2+r)$, $f(0,-r)=r(-2+r,-3)$, and $f(-r,0)=r(-3,-2+r)$, which is symmetric with respect to a line $x=y$ and a point $f(0,0)=(0,0)$ is contained in the bounded component of $\Bbb R^2\setminus f(B)$.
Let $G$ be a geometric graph with vertices $(0,0)$, $(0,r)$, $(r,0)$, $(0,-r)$, and (eight) edges connecting them induced from $B$ and diagonals of $S$. It is easy to check that $G$ is three-connected.
Let $L$ be any of latin numbers from $I$ to $IV$ and $f_L$ be the restriction of $f$ onto $L$-th (closed) quadrant $Q_L$ of $\Bbb R^2$. Then $f_L$ coincides with the restriction onto $L$ of $f_+$ (if $L=I$ or $L=III$) or of $f_-$ (if $L=II$ or $L=IV$).
Let $B^*$ be the “support” of $G$, that is a union of $B$ with the diagonals. We claim that $f|G$ is injective. Indeed, let $z$ and $z’$ be distinct points of $B^*$. If $z, z’\in B$ then $f(z)\ne f(z’)$, because $f|B$ is injective. If $z\in B^*\setminus B$ then $z$ belongs to one of diagonals of $S$. Then $z\in (Q_I\cup Q_{III})\cap (Q_{II}\cup Q_{IV})$, so $f(z)=f_+(z)=f_-(z)$. If $z’\in Q_I\cup Q_{III}$ then $f(z’)=f_+(z’)\ne f_+(z)=f(z)$. If $z’\in Q_{II}\cup Q_{IV}$ then $f(z’)=f_-(z’)\ne f_-(z)=f(z)$.
Therefore $f$ induces a plane embedding $\hat f$ of $G$.  By Whitney's theorem, all plane embeddings of a three-connected planar graph are equivalent, that is, obtainable from one another by a plane homeomorphism up to the choice of outer face. Since $f(0,0)=(0,0)$ is contained in the bounded component of $\Bbb R^2\setminus f(B)$, both  $G$ and $\hat f(G)$ have as the outer face the unique face with four vertices. Therefore there exists a homeomorphism $h$ of $\Bbb R^2$ such that $hf|B^*$ is an identity map.
Let $L$ be any of Latin numbers from $I$ to $IV$. Since a triangle $\Delta_L=S\cap Q_L$ is compact and the function $hf_L$ is continuous and injective, $hf_L|\Delta_L$ is a homeomorphic embedding.  Let $B_L$ be the boundary of  $\Delta_L$. Since a boundary of a compact subset $K$ of $\Bbb R^2$ consists of points $x$ of $K$ having no neighborhood in $K$ homeomorphic to an open disk, $hf_L(B_L)=B_L$ is the boundary of  $hf_L(\Delta_L)$. The set $\Bbb R^2\setminus B_L$  consists of two connected components  $\Delta_L\setminus B_L$ and $\Bbb R^2\setminus\Delta L$, so a connected set $C=hf_L(\Delta_L\setminus B_L)$ is contained in one of them. It is easy to construct a continuous map $r$ from a set $(\Bbb R^2\setminus\Delta L) \cup B_L$ onto its boundary $B_L$ such that $r|B_L $ is the identity map, that is, $r$ is a retraction. If $C\subset \Bbb R^2\setminus\Delta_L$ then $r|hf_L(\Delta_L)$ is a retraction of the set $ hf_L(\Delta_L)$ onto its boundary $B_L$, which contradicts the drum theorem, since $ hf_L(\Delta_L)$ is homeomorphic to a closed disk. Therefore $C=hf_L(\Delta_L\setminus B_L)\subset \Delta_L\setminus B_L$. Since this holds for each $L$, we have that the map $hf_S$ is injective, and so is $f_S$. $\square$
