Injectivity of the Differential of Smooth Map I am trying to answer the following question:
Let $M = \{(x,y)\in \mathbf{R}^2 : x^2 + y^2 < 1\}$. Define a smooth or $C^\infty$ function by $f\colon M \rightarrow \mathbf{R}^2$ as $$(x,y)\rightarrow  \left(\frac{y}{1-x^2-y^2}, e^{x^2} \right).$$ 
a) Find the set $S$ of points $p$ of at which of $f_*{p}$ (the differential) is injective.
b) Prove that $f(S)$ is an open subset of $\mathbf{R}^2$.
For part a), I have done the following:
The differential of the mapping is given by the Jacobian matrix at an arbitrary point p.
I.E.:
$$
f_*{p}= 
\left(
\begin{matrix} \frac{2xy}{(1-x^2-y^2)^2}&\frac{1-x^2+y^2}{(1-x^2-y^2)^2}\\ 2xe^{x^2}&0\\
\end{matrix}
\right)
$$
Since $\operatorname{rank}(f_*{p}) < 2$, from the above then $f$ fails to be an immersion. Since the Jacobian is $2\times 2$, we can just take the determinant and check when it equals zero and by the I.F.T (the Inverse function Theorem), we see that:
$\det (f_*{p}) = -2xe^{x^2}\frac{1-x^2+y^2}{(1-x^2-y^2)^2}$ , and when this equals zero, you would get $x=0$, or $y=\pm 1$. To then conclude what set $S$ for which the differential is injective, then I conclude that $S = M \setminus \{(0,y): y \in  (-1,1)\}$. I hope that this is enough...
b) I am a bit stuck on this one, to be honest. First I try to find f{(0,y): y$\epsilon$ (-1,1)}, which is part of what I found in part a). For that, I would get: {($\frac{y}{(1-y^2)^2}$,1)}=(-$\infty$,$\infty$)x{1}, which is itself a subset of $R^2$. I am then trying to apply the following logic: f(S) = f(M - (-$\infty$,$\infty$)x{1})$\supset$f(M)-f($\infty$,$\infty$)x{1}), but am not sure how this even helps.
Any feedback, particularly on part b), is much appreciated.
Thanks!

To complete the argument above, one can use the observation that:
$f(M)= ${$ (x,y)\epsilon$$R^2 1$$\leq$$y$$<$e }.
it is clear then that $f(S) =${$ (x,y)\epsilon$$R^2 1$$<$$y$$<$e }
which is an open set in $R^2$.
