What is the inverse of the parabolic mapping given by $F(\sigma, \tau) = (\sigma\tau, (\tau^2-\sigma^2)/2) $ 
Given the mapping
  $$F:\mathbb{R}^2\to\mathbb{R}^2;\ \ \ F(\sigma,\tau) = (\sigma\tau, (\tau^2 - \sigma^2)/2) = (x,y) \text{(Cartesian)},$$
  We can use the fact that $F$ is invertible where $(\sigma, \tau)\neq (0,0)$, and the image of this domain under $F$ is $\mathbb{R}\backslash\{x|x\neq 0\}$, and $x^2+y^2 = (\tau^2 + \sigma^2)^2/4$. We must find the inverse mapping.

I know from the Invertible Function Theorem that if $DF$ is the partial derivative matrix, then $DF^{-1}(x, y) = [DF(\sigma,\tau)]^{-1}$.
So,
$$DF(\sigma,\tau) = \begin{pmatrix} \tau & \sigma \\ -\sigma & \tau \end{pmatrix}$$
Therefore,
$$ DF^{-1}(x,y) = \frac{1}{\tau^2 + \sigma^2} \begin{pmatrix} \tau & -\sigma \\ \sigma & \tau \end{pmatrix}$$

Here is where I start to get uncomfortable. I need to change $\tau$ and $\sigma$ to $x$ and $y$ for this to be valid, and to use partial integration to work backwards from the inverse derivative matrix in order to find $F^{-1}$. This is going to be messy and unpleasant. It would be a little nicer if I can just use the matrix in terms of $\sigma$ and $\tau$, but since I know that $F$ take a point in $\sigma, \tau$ space to $x, y$ space, the inverse should do the opposite, and I want the inverse partial derivative matrix to have $x$ and $y$ in it. Here is my attempt with that:

$x = \sigma\tau$, so $\sigma = x/\tau$.
$y=(\tau^2-(s^2/\tau^2))/2\Rightarrow \tau^4-2y\tau^2-x^2 = 0$, so
$$\tau = \pm \sqrt{y + \sqrt{x^2+y^2}},\ \ \ \ \sigma = \frac{x}{\sqrt{y + \sqrt{x^2+y^2}}}$$
So if $F^{-1}(x,y) = (g(x,y), h(x,y))$,
$$ DF^{-1}(x,y) = \frac{2}{\sqrt{x^2+y^2}} 
\begin{pmatrix} 
\pm \sqrt{y + \sqrt{x^2+y^2}} & \mp\frac{x}{\sqrt{y + \sqrt{x^2+y^2}}} \\ 
\pm\frac{x}{\sqrt{y + \sqrt{x^2+y^2}}} & \pm \sqrt{y + \sqrt{x^2+y^2}} 
\end{pmatrix}
=
\begin{pmatrix}
g_x & g_y\\
h_x & h_y
\end{pmatrix}$$
This is beyond what I think would be required to integrate for this course, so I'm really struggling to find a nicer alternative. However, using WolframAlpha to do the grunt work for me, I find
$$\begin{aligned}
g_x(x,y) = \pm\frac{4x}{\sqrt{\sqrt{x^2+y^2}+y}} & g_y(x,y) = \text{some arctangent stuff}\\
h_x(x,y) = \text{No exact solution} & h_y(x,y) = \pm 4\sqrt{\sqrt{x^2+y^2}+y}
\end{aligned}$$
So clearly I've lost something, but I'm not sure what.
 A: Prologue. Here is a way to find the inverse of $F$ without dealing with the Jacobian matrix.  It is important to note that $F$ is not invertible.    It only has local inverses around certain points of $\mathbb{R}^2$.  Note that this approach is only applicable to $F$ due to its symmetry.  In general, analyzing the Joacobian matrix to identify the points at which the function is locally invertible and then applying the Inverse Function Theorem is probably a better approach.  
First, note that $F(-s,-t)=F(+s,+t)$ for all $s,t\in\mathbb{R}$.  Therefore, $F$ is not injective on any open neighborhood of $(0,0)$.  Thus, $F$ has no local inverses around $(0,0)$.  Define $S_+$, $S_-$, $T_+$, and $T_-$ for the subsets of $\mathbb{R}^2$ of the form
$$S_+:=\big\{(s,t)\in\mathbb{R}^2\,|\,s>0\big\}\,,\,\,S_-:=\big\{(s,t)\in\mathbb{R}^2\,|\,s<0\big\}\,,$$
$$T_+:=\big\{(s,t)\in\mathbb{R}^2\,|\,t>0\big\}\,,\,\,T_-:=\big\{(s,t)\in\mathbb{R}^2\,|\,t<0\big\}\,.$$
Write $V:=\mathbb{R}^2\setminus\big\{(0,0)\big\}$.  We shall exhibit the local inverses $F_1^{-1}:V\to S_+$, $F_2^{-1}:V\to S_-$, $F_3^{-1}:V\to T_+$, and $F_4^{-1}:V\to T_-$ of $F$.
First, each $(s,t)\in V$ lies in one of $S_+$, $S_-$, $T_+$, and $T_-$.  Suppose that $F(s,t)=(x,y)$.  Then,
$$st=x\text{ and }\frac{t^2-s^2}{2}=y\,.$$
Define 
$$P(\lambda):=\lambda^2+2y\,\lambda-x^2=\lambda^2-(s^2-t^2)\,\lambda-s^2t^2\,.$$
Thus, $$P(\lambda)=(\lambda-s^2)(\lambda+t^2)\,.$$  The roots of $P(\lambda)$ are then $\lambda=s^2$ and $\lambda=-t^2$.  Using the quadratic formula, the roots of $P(\lambda)$ are $-y\pm\sqrt{x^2+y^2}$.  Since $s^2\geq 0$ and $-t^2\leq 0$, we must have $$s^2=-y+\sqrt{x^2+y^2}\text{ and }-t^2=-y-\sqrt{x^2+y^2}\,,$$ so that
$$s=\pm\sqrt{\sqrt{x^2+y^2}-y}\text{ and }t=\pm\sqrt{\sqrt{x^2+y^2}+y}\,.$$
If $(s,t)\in S_+$, then we can choose the local inverse $F_1^{-1}:V\to S_+$ to be
$$F_1^{-1}(x,y):=\Biggl(+\sqrt{\sqrt{x^2+y^2}-y},+\text{sign}(x)\,\sqrt{\sqrt{x^2+y^2}+y}\Biggr)$$
for all $x,y\in V$.  If $(s,t)\in S_-$, then we can choose the local inverse $F_2^{-1}:V\to S_-$ to be
$$F_2^{-1}(x,y):=\Biggl(-\sqrt{\sqrt{x^2+y^2}-y},-\text{sign}(x)\,\sqrt{\sqrt{x^2+y^2}+y}\Biggr)\,.$$
If $(s,t)\in T_+$, then we can choose the local inverse $F_3^{-1}:V\to T_+$ to be
$$F_3^{-1}(x,y):=\Biggl(+\text{sign}(x)\,\sqrt{\sqrt{x^2+y^2}-y},+\sqrt{\sqrt{x^2+y^2}+y}\Biggr)\,.$$
If $(s,t)\in T_-$, then we can choose the local inverse $F_3^{-1}:V\to T_-$ to be
$$F_4^{-1}(x,y):=\Biggl(-\text{sign}(x)\,\sqrt{\sqrt{x^2+y^2}-y},-\sqrt{\sqrt{x^2+y^2}+y}\Biggr)\,.$$
Epilogue.  For $(s,t)\in\mathbb{R}^2$ and $(s',t')\in\mathbb{R}^2$, observe that $F(s,t)=F(s',t')$ if and only if $(s',t')=\pm (s,t)$.  Therefore, $F^{-1}$ has two branches and $(0,0)$ is its unique branch point.  
By identifying the complex plane $\mathbb{C}$ with $\mathbb{R}^2$ via $z\mapsto \big(\text{Im}(z),\text{Re}(z)\big)$ for every $z\in\mathbb{C}$, we actually see that
$$F(z)=\frac{1}{2}\,z^2$$
for all $z\in\mathbb{C}$.  The task is basically calculating the square-root functions of the complex plane.
