<s>[INST] Find the derivative of the inverse function, $Dg(0,1)$. Let $f:\mathbb{R}^2\to \mathbb{R}^2$ be defined by the equation
\begin{equation} f(x,y)=(x^2-y^2,2xy)\end{equation}
Parts (a) and (b) of the problem asked me to show that $f$ is one-to-one on the set $A$ consisting of all $(x,y)$ with $x>0$, and to find set $B=f(A)$, which was easy enough to do. Part (c) asks

If $g$ is the inverse function, find $Dg(0,1)$.

What I thought to do was to find the matrix $Df$ and find the inverse of it, but I am finding that very challenging to do (here I thought I had my linear algebra skills in check!) Because what I get is that
$Df=
\begin{bmatrix}
2x &2y\\
-2y & 2x
\end{bmatrix}$
And when I try to find the inverse, i get to the step
$
\begin{bmatrix}
x & y & 1/2 & 0\\
0 & \frac{x^2+y^2}{x} & \frac{y}{2x} & 1/2
\end{bmatrix}$
And i feel like maybe this is not the correct method to finding the inverse...?
Thank you!
 [/INST] First, fix the differential ($f(x,y)=(u(x,y),v(x,y))\;;u=x^2-y^2\;;v=2xy$):
$$Df=
\begin{pmatrix}
\partial u/\partial x &\partial u/\partial y\\
\partial v/\partial x & \partial v/\partial y
\end{pmatrix}=
\begin{pmatrix}
2x &-2y\\
2y & 2x
\end{pmatrix}$$
Now, with the invaluable aid of the commentators, the inverse is
$$Dg(f(x,y))=\dfrac{1}{\vert Df\vert}\begin{pmatrix}2x &2y\\ -2y & 2x\end{pmatrix}=\dfrac{1}{4(x^2+y^2)}\begin{pmatrix}2x &2y\\ -2y & 2x\end{pmatrix}=$$
$$=\dfrac{1}{2(x^2+y^2)}\begin{pmatrix}x &y\\ -y & x\end{pmatrix}$$
We have to find the values for wich $f(x,y)=(0,1)$, so is, $x^2-y^2=0$ and $2xy=1\implies (x,y)=(1/\sqrt{2},1/\sqrt{2})$ or $(x,y)=(-1/\sqrt{2},-1/\sqrt{2})$. But $x\gt0$, so we drop the solution with the negatives. Then,
$$Dg(0,1)=Dg(f(1/\sqrt{2},1/\sqrt{2}))=\dfrac{1}{2}\begin{pmatrix}\sqrt{2}/2 &\sqrt{2}/2\\-\sqrt{2}/2 &\sqrt{2}/2\end{pmatrix}=$$
$$Dg(0,1)=\dfrac{1}{4}\begin{pmatrix}\sqrt{2} &\sqrt{2}\\-\sqrt{2} & \sqrt{2}\end{pmatrix}$$
</s>