Solution: We will use the inverse function theorem to find the Jacobian matrix of $g$ at $f(2,1)$. Since $g$ is the inverse of $f$, we have \[ Dg(f(x,y)) = \left( Df(x,y) \right) ^{-1}, \] where $Dg$ and $Df$ are the derivative matrix of $g$ and $f$ respectively. Thus, \begin{align*} Dg(f(2,1)) & = \begin{bmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\[2ex] \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \\ \end{bmatrix}_{(2,1)}^{-1} \\[2ex] & = \begin{bmatrix} 2x & -2y \\ 2y & 2x \\ \end{bmatrix}_{(2,1)}^{-1} = \begin{bmatrix} 4 & -2 \\ 2 & 4 \\ \end{bmatrix}^{-1} . \end{align*} Since $\det A^{-1} = (\det A)^{-1} $, we have \begin{align*} \det \begin{bmatrix} 4 & -2 \\ 2 & 4 \\ \end{bmatrix}^{-1} = \frac{1}{20}. \end{align*} Hence, the determinant of the Jacobian matrix of $g$ at $f(2,1)$ is equal to $\frac{1}{20} = 0.05$.