A radius for which the inverse of a function is well defined (Inverse Function Theorem) Let $f(x,y)=(\frac{1}{2} x^2 + x(y-1)^3, xy-x)$
(1) Find a formula for $f^{-1}$ in a small ball $B(b,r)$ where $b=(\frac{1}{2}, 0)$.
(2) Give an example of a radius $r>0$ for which the inverse function is well defined.

I showed that the inverse was well defined in some neighbourhood of b in a previous question using the Inverse Function Theorem.
 A: The advantage of not being Mathematica is that I can use simpler symbols.
Suppose that our function maps $(x,y)$ to $(s,t)$.  We want to recover $x$ and $y$ given $s$ and $t$.  It turns out that we don't really need Mathematica.
We know that
$$\frac{1}{2}x^2+x(y-1)^3=s \qquad\text{and}\qquad xy-x=t.$$
The second equation looks more tractable.  Trusting that $x=0$ is not going to give us a problem, we rewrite the equation as
$$y-1 =\frac{t}{x}.$$
Substituting in the first equation, we see that perhaps we are doing something right, since we get
$$\frac{1}{2}x^2 + \frac{t^3}{x^2}=s.$$
Solve for $x$.  The rest will be straightforward, since if we have $x$ in terms of $s$ and $t$, then from
$$y=1 +\frac{t}{x},$$
we also have $y$ in terms of $s$ and $t$.
Rewrite the equation for $x$ as
$$x^4 -2sx^2 +2t^3=0.$$
This is a quadratic in $x^2$, readily solved for $x^2$, and hence for $x$.  We get
$$x=\pm\sqrt{s\pm \sqrt{s^2-2t^3}}.$$
A: I will cheat. The command in Mathematica is:
Solve[{f1, f2} == {x^2/2 + x (y - 1)^3, x y - x}, {x, y}]

The result is:

{x -> (1/(2 f2))((2 Sqrt[2] f1 Sqrt[-f2 (-f1 + Sqrt[f1^2 - 2 f2^3])])/f2 - (Sqrt[2] (-f2 (-f1 + Sqrt[f1^2 - 2 f2^3]))^(3/2))/f2^2),y -> (2 f2 + Sqrt[2] Sqrt[f1 f2 ->f2 Sqrt[f1^2 - 2 f2^3]])/(2 f2)}

and similar solutions with different signs. You have to choose the right signs. They're just playing "I Saw the Sign" on radio. ;-) All of the four solutions returned by Mathematica look pretty messy but one of them is the right one.
