Q:

Determine whether f is a bijection in neighbourhood of singular points

Given a function $f:\mathbb{R}^2\to\mathbb{R}^2$, $f(x,y)=(x^2+y-y^2, 2xy+y)$, determine for which points $(x_0,y_0)$ where $JDf(x_0,y_0)=0$ function $f$ is bijective from some open set containing $(x_0,y_0)$ to an open set containing $f(x_0,y_0)$.
My approach:
We have $Df(x,y)=
\begin{vmatrix}
2x & 1-2y\\
2y & 2x-1
\end{vmatrix}
= 4x^2+2x + 4y^2-2y=0 \iff (x+\frac{1}{4})^2 + (y-\frac{1}{4})^2 = \frac{1}{8}$. Now, $f$ is continous, thus an inverse image of every open set is an open set. Hence $f$ is bijective around some point (x,y) $\iff$ $f$ is injective. Letting $\begin{cases}  u =x^2+y-y^2\\v=2xy+y\end{cases}$ and for $x\neq -\frac{1}{2}$ and $y\neq 0$ we obtain equations for $x$ and $y$:
$u=x^2+\frac{v}{2x+1}-(\frac{v}{2x+1})^2$
$u=(\frac{v-y}{2y})^2+y-y^2$, but I don't know how to show how many solutions they have around points of that circle.
On the other hand it is very easy to show that for $x=0$ and $y=0$ $f$ is not bijective. It suffice to examine $f$ on a line $(x,0)$- we have $v=0$ and $u=x^2$, thus it's not an injection. Similar thing works for $(-\frac{1}{2}, \frac{1}{2})$, but again- I'm not sure how to deal with $(-\frac{1}{2},0)$ and $(0,\frac{1}{2})$.
My additional question is: how to deal in general with such problems when we cannot use the IFT? That is what I've found. Any other suggestions? 

A:

Specific example
When $f$ is a quadratic polynomial, $\det Df(a)=0$ implies the failure of local injectivity at $a$. Indeed, the Taylor formula takes the form
$$
f(a+h) = f(a) +   Df(a) h  + \frac12 h^T D^2f(a)h
$$
Consider $h\in \ker Df(a)$: the right hand side is an even function of $h$.
General remarks
Investigation of injectivity in the vicinity of a point $a$ with $\det Df(a)=0$ is tricky. If $\det Df$ takes on both positive and negative values in every neighborhood of $a$ (which is the case in your example), local injectivity fails. This requires a topological argument based on degree: a homeomorphism of $U$ is either topologically sense-preserving or topologically sense-reversing everywhere in $U$, and at every point with $\det Df\ne 0$ the sign of $\det Df$ must be consistent with this topological  behavior. See, e.g., Lemma 5.10 in Degree Theory in Analysis and Applications by Fonseca and Gangbo. 
If in some neighborhood the inequality $\det Df\ge 0$ holds, then local injectivity may hold or fail; it's really hard to tell without looking at the specific map. Two examples with nonnegative Jacobian determinant are 
$$(x,y)\mapsto (x^3+xy^2, x^2y+y^3)\tag{globally injective}$$
$$(x,y)\mapsto (x^2-y^2, 2xy)\tag{not injective near (0,0)}$$

