Is the determinant of the Jacobian matrix of $g$ at $f(2,1)$ correct?

Let $$f(x,y)=(x^2-y^2,2xy),$$ where $$x>0,y>0.$$ Let $$g$$ be the inverse of $$f$$ in a neighbourhood of $$f(2,1)$$. Then the determinant of the Jacobian matrix of $$g$$ at $$f(2,1)$$ is equal to...............?

Solution: Let $$f(x,y)=(f_1(x,y),f_2(x,y))$$,where $$f_1(x,y)=x^2-y^2$$ and $$f_2(x,y)=2xy$$,then the Jocobian of the matrix is,$$Jf=\begin{bmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \end{bmatrix}=\begin{bmatrix} 2x & -2y \\ 2y & 2x \end{bmatrix}$$,then $$Jf(2,1)=\begin{bmatrix} 4 & -2 \\ 2 & 4 \end{bmatrix}$$,$$det(Jf(2,1))=det \begin{bmatrix} 4 & -2 \\ 2 & 4 \end{bmatrix}=20\neq 0$$,then by Inverse function theorem $$det (Jg(2,1))=det (Jf^{-1}(2,1))=\frac{1}{det((Jf(2,1)))}=1/20$$

Is my Solution correct?Please suggest anyother method by which it can be tackled also check whether I applied the Inverse function theorem correctly

$$f(2,1)=(3,4)$$ so you want $$\det(\mathrm J_g(3,4)) = \det(\mathrm J_{f^{-1}}(3,4))$$
But otherwise, yes, that is: $$\det(\mathrm J_f(2,1))^{-1}=1/20$$.