Let $f(x,y)=(x^2-y^2,2xy)$
Prove for all point $(x_0,y_0)\not =(0,0)$ the restriction of $f$ to neighborhood $(x_0,y_0)$ have inverse.
My work:
Let $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ defined by $f(x,y)=(x^2-y^2,2xy)$
Let $(x_0,y_0)\in \mathbb{R}^2$ with $(x_0,y_0)\not =(0,0)$
let go to see $f'(x_0,y_0)$ is invertible.
$$f'(x_0,y_0)=\begin{pmatrix} \nabla f_1(x_0,y_o) \\ \nabla f_2(x_0,y_0) \end{pmatrix}=\begin{pmatrix} 2x_0 &-2y_0\\ 2y_0 & 2x_0 \end{pmatrix}$$
Then $\det(f'(x_0,y_0))=4(x_0^2+y_0^2)\not =0$ this implies$f'(x_0,y_0))$ is invertible.
Let $U=?$ a neighborhood of $(x_0,y_0)$. (By theorem of inverse function the exercise can be solved but i'm stuck defining the neighborhood $U$)
I am stuck here. Can someoe give me a hint?
Suppose w.l.o.g. that $\;x_0\neq0\;$ , then take $\;r:=\frac{|x_0|}2\;$ , and thus you can take
$$U:=B_{(x_0.y_0)}(r):=\text{ the open ball with center$\,(x_0,y_0)\,$ and radius}\;r$$