How to use the inverse function theorem? I have a function $F (x, y) = (x^2+y^2, xy)$ and I need to show that it has an inverse. How do I find the inverse of this function using the inverse function theorem? I have not learned this before in all of my math classes and I just need some brushing up on how to use it.
Thank you for your assistance.
edit: On the set $\{(x, y) : −x < y < x\}$.
 A: For $F(x,\, y) = (x^2+y^2,\; xy)$ denote
$$\begin{cases}u = x^2+y^2, \\
v = xy \end{cases}$$
Then
$$\begin{cases}
u+2v=(x+y)^2, \\
u-2v=(x-y)^2 
\end{cases} $$
On the given set $\{(x, y):\;\; −x < y < x \} $ we have $$ x+y>0,\;\;x-y>0, \quad J_F(x,\, y)=2(x^2-y^2)\ne{0},$$ therefore $F$ is invertible and 
$$
\begin{cases}
x+y=\sqrt{u+2v}, \\
x-y=\sqrt{u-2v} 
\end{cases}$$
Thus
$$\begin{cases}
2x=\sqrt{u+2v}+\sqrt{u-2v},\\
2y=\sqrt{u+2v}-\sqrt{u-2v}.
\end{cases}$$
A: The function
$$F:\quad  \Omega\to{\mathbb R}^2,\qquad  (x,y)\mapsto (u,v):=(x^2+y^2,xy)$$
has an inverse for suitably chosen $\Omega\subset{\mathbb R}^2$ not because of the inverse function theorem, but because you are able to compute this inverse explicitly, as has been done in M. Strochyk's answer. Note that to arrive at the formulae
$$F^{-1}:\quad(u,v)\mapsto (x,y)=\left({\sqrt{u+2v}+\sqrt{u-2v}\over2},{\sqrt{u+2v}-\sqrt{u-2v}\over2}\right)$$
no checking of Jacobians was necessary. In addition we have a clear picture of the domain of $F^{-1}$ corresponding to the given $\Omega$.
The inverse function theorem is needed for theoretical considerations, and in cases where the inverse $F^{-1}$ cannot be expressed in terms of known functions. It is a purely local theorem: Given a point $(x_0,y_0)$ with $F(x_0,y_0)=:(u_0,v_0)$ it guarantees the existence of a "window" $U$ with center $(x_0,y_0)$ such that the restriction $F_{loc}:=F\restriction U$ has an inverse $F^{-1}_{loc}:\ F(U)\to U$ which is again differentiable. The essential technical condition is the nonvanishing of the Jacobian of $F$ at $(x_0,y_0)$. The theorem doesn't tell you what the "maximal domain" of this local inverse could be.
A: 
when you calculate jacobian matrix and found value p,where this matrix's determinant is not zero,there maybe you could you this formula. i would be happy if it helps you.also  you could check  all formula in wikipedia's site
.for single variable .let suppose that
$y=\sqrt{x}$
and   let us replace  places for x and y,or we get
$x=\sqrt{y}$
from which
$y=x^2$
