Inverse Function Theorem and finding local inverses

Question

$f(x,y) = (x+y,x^2+y)$ for this function, the question is Write down a $C^1$ local inverse around (0, 0).
I wonder how can I find the points f has a local inverse? I am trying to use Lagrange multipliers but I am confused, and also how can I understand it does not have a local inverse for any point for example (1/2,0).
I am a new learner for this topic and some notions are really interesting for me, and this type of question is hard to think for me.

Answer 1

We have the theorem: Let $f$ is a function from open set in $\mathbb{R}^{n}$ to $\mathbb{R}^{n}$ and $f$ is of class $C^r$. If $Df(x)$ (the derivative) is non-singular at the point $\textbf{a}$ of $A$, then there is a neighborhood $U$ of the point $\textbf{a}$ such that $f$ is one-to-one from $U$ to an open set $V$ of $\mathbb{R}^{n}$ (hence the inverse function exists), and the inverse function is of class $C^{r}$.

So to determine that $f$ has local inverse in your question, it is sufficient to check the singularity of the $Df$ at the given point.

Answer 2

The inverse is a mapping from $(u,v)$ to $(x,y)$, with $g(u,v)=(x,y)$, so that

$$x+y=u \\ x^2+y=v.$$

This system can be solved algebraically, but there are two solutions unless $u-v \geq 1/4$:

$$x^2+y-(x+y)=v-u \\ x=\frac{1 \pm \sqrt{1-4(u-v)}}{2} \\ y=u-x=\frac{2u-1 \mp \sqrt{1-4(u-v)}}{2}.$$

However, because $f(0,0)=(0,0)$, you want $g(0,0)=(0,0)$. Only one of these two mappings has this property.