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.
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.