Proving Injectivity The problem is to show the function $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$ given by
$$f(x,y)=(\tfrac{1}{2}x^2+y^2+2y,\,x^2-2x+y^3)$$ is injective on the set
$$M=\{(x,y)\in\mathbb{R}^2:|x-1|+|y+1|<\tfrac{1}{6}\}.$$
My idea is to consider the following map (here, $u,v\in\mathbb{R}^2$):
$$\phi_v(u)=u-f(u)+v,\quad v\in M$$
If I manage to show that


*

*$\phi_v:D\rightarrow D$ is well-defined for some closed sets

*$\phi_v$ is a contraction (Lipschitz constant $<1$) on $D$
then by the Contraction Mapping Theorem, $\phi_v$ has a unique fixed point. Hence, $v$ has a unique preimage $u$ for each $v$. i.e. $f$ is injective as desired.
But I ran into troubles when I attempted to find a suitable closed set $D$. Obviously it depends on the domain $M$. $M$ given here is really weird so I am not too sure how to proceed.
 A: I'll present an approach along the lines of my comment. First, I'll normalize the derivatives at $(1,-1)$ by dividing the second component by $3$: 
$$\tilde f(x,y) = (x^2/2+y^2+2y, (x^2-2x+y^3)/3)$$
This is done so that the Jacobian matrix of $\tilde f$ at $(1,-1)$ is the identity. Now split $\tilde f=L+g$ where $L(x,y)=(x-3/2,y+1/3)$ is the linear part and $g$ is the rest. If we can show that $g$ is Lipschitz with a constant less than 1, we are done. 
The first component of $g$ is $g_1(x,y)=x^2/2+y^2+2y-x+3/2$, with the gradient $\nabla g_1=\langle (x-1),2(y+1)\rangle$. We estimate the gradient by $|\nabla g_1|< \sqrt{5}/6<1/2$. 
The second component of $g$ is $g_2(x,y)= (x^2-2x+y^3-3y-1)/3$, with the gradient $\nabla g_2=\langle 2(x-1)/3,y^2-1\rangle$. Since $|y^2-1|\le (|y+1|+2)|y+1|<13/36$, we obtain  $|\nabla g_2|\le \sqrt{1/81+(13/36)^2}<1/2$. 
Since both components have Lipschitz constant $<1/2$, the map $g$ has Lipschitz constant $<1$. (Of course a more precise bound can be given, but this suffices.)
A: $\newcommand{\ve}{\varepsilon}\newcommand{\abs}[1]{|{#1}|}$
I've tried a different approach, a few times I made a mistake before I got this version, so I don't guarantee that there's not still one more error which I did not notice. 
We have function $f(x,y)=(f_1(x,y),f_2(x,y))$ where
$$
\begin{align}
f_1(x,y)&=\frac{1}{2}x^2+y^2+2y;\\
f_2(x,y)&=x^2-2x+y^3.
\end{align}
$$
We want to find out whether this function is injective on a small region around the point $x=1$, $y=-1$.
Let us try to use directly the definition of injective function. Let's have a look what can be said if $f(x_1,y_1)=f(x_2,y_2)$.
Notice that 
$$\begin{align}
f_2(x,y)-2f_1(x,y)&=y^3-2y^2-4y-2x\\
2f_1(x,y)+2&=x^2+2(y+1)^2
\end{align}$$
So we have
$$
\begin{align}
y_1^3-2y_1^2-4y_1-2x_1&=y_2^3-2y_2^2-4y_2-2x_2\\
x_1^2+2(y_1+1)^2&=x_2^2+2(y_2+1)^2
\end{align}
$$
which gives
$$
\begin{align}
2(x_1-x_2)&=(y_1^3-2y_1^2-4y_1)-(y_2^3-2y_2^2-4y_2) \\
(x_1-x_2)(x_1+x_2)&=2[(y_2+1)^2-(y_1+1)^2]
\end{align}
$$
Now we will try to use that $\abs{y+1}<\ve$ and $\abs{x-1}<\ve$.
For the function $g(y)=y^3-2y^2-4y$ we have $g'(y)=3y^2-4y-4=3(y+1)^2-10(y+1)+3$. Hence $\abs{g'(y)-3}\le 10\ve+3\ve^2$ and $\abs{g'(y)}\le 3+11\ve$ if $\ve$ is small enough. By mean value theorem we get
$$2\abs{x_1-x_2} \le \abs{y_1-y_2}(3+11\ve)$$
We also have
$\abs{x_1+x_2}\le \abs{x_1}+\abs{x_2}\le 2(1+\ve)$, which gives
$$\abs{(x_1-x_2)(x_1+x_2)}\le (3+11\ve)(1+\ve)\abs{y_1-y_2}.$$
Again, for small enough $\ve$, we get
$$\abs{(x_1-x_2)(x_1+x_2)}\le (3+16\ve)\abs{y_1-y_2}.$$
On the other hand, using mean value theorem for the function $y\mapsto (y+1)^2$ we can find out that 
$$\abs{(y_2+1)^2-(y_1+1)^2} \ge 2(2-\ve)\abs{y_2-y_1}$$
If both these estimates are true for $\abs{y_2-y_1}\ne0$ than
$$3+16\ve\ge 8-4\ve$$
which gives $\ve\ge\frac5{20}=\frac14$. 
So for $\ve=\frac16$ this cannot be true. (I believe $\ve\le\frac16$ was sufficient in all estimates above, but you should double check this.)
