Derivative of the Inverse of a multivariable real function

Question

I have the function $$ f(x,y) = (x^2-y^2, 2xy),\,\,\,x>0 $$ Let $g=f^{-1}$. The goal is to compute $Dg(0,1)$. The function is known to be 1-1 (hence the existence of $g$). So we have $$ Dg(0,1) = [Df(g(0,1))]^{-1} $$ My problem is that I don't want to compute $g$ (in order to evaluate $g(0,1)$), is there way of doing so?

Answer 1

Note that $f\left(\frac1{\sqrt2},\frac1{\sqrt2}\right)=(0,1)$, and since the Jacobian of $f$ at $\left(\frac1{\sqrt2},\frac1{\sqrt2}\right)$ is$$\begin{bmatrix}\sqrt2&-\sqrt2\\\sqrt2&\sqrt2\end{bmatrix},$$the Jacobian of $g(=f^{-1})$ at $(0,1)$ is$$\begin{bmatrix}\sqrt2&-\sqrt2\\\sqrt2&\sqrt2\end{bmatrix}^{-1}=\begin{bmatrix}\frac1{2\sqrt{2}}&\frac1{2\sqrt{2}}\\-\frac1{2\sqrt{2}}&\frac1{2\sqrt{2}}\end{bmatrix}.$$So,$$Dg(0,1)(x,y)=\left(\frac{x+y}{2\sqrt2},\frac{-x+y}{2\sqrt2}\right).$$