$\newcommand{\ve}{\varepsilon}\newcommand{\abs}[1]{|{#1}|}$ 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.)