Is function invertible? Reflection on the unit circle:
Let $E=\mathbb R ^{2} - \left\{0,0\right\} $ be perforated plane and $f: E \mapsto E$ defined by $f\left(x,y\right)=\left(\frac{x}{x^{2}+y^{2}  }  , \frac{y}{x^{2}+y^{2}  }   \right) 
$
Show with Jacobian matrix that $f$ is in all points local invertible. Show that $f$ is also global invertible. Find $f^{-1}$ and explain mapping geometrically.
What I did:
I started with determinante of Jacobian matrix to show that function is local invertible, but I got the following result. 
Is that enough for showing that function is local invertible? How do I do rest?
$Df=\frac{-x^{4}-y^{4}-2x^{2}y^{2}    }{\left(x^{2}+y^{2}  \right) ^{2} } $

See the comment on the question for local invertibility. 
Now, we prove that $f$ is invertible by showing that $f$ is one-one and onto. Given $(x_1, y_1), (x_2,y_2) \in \Bbb R^2 - \{(0,0)\}$, we have:
$$f(x_1,y_1) = f(x_2, y_2) \implies \begin{cases} \frac{x_1}{x_1 ^2 + y_1^2} = \frac{x_2}{x_2 ^2 + y_2^2} \\ \frac{y_1}{x_1 ^2 + y_1^2} = \frac{y_2}{x_2 ^2 + y_2^2}\end{cases}$$
Squaring the equations and adding them to each other, we get $x_1^2 + y_1^2 = x_2^2 + y_2^2$, which then given $x_1 = x_2$ and $y_1 = y_2$.
Next, given $(X,Y) \in \Bbb R^2 - \{(0,0)\}$, we want to find $(x,y)$ such that $f(x,y) = (X,Y)$, so we are solving for $x$ and $y$:
$$\begin{cases} X = \frac{x}{x^2 + y^2} \\ Y = \frac{y}{x^2 + y^2} \end{cases}$$
Notice then that $X^2 + Y^2 = \frac1{x^2 + y^2}$
Edit
Now, on one hand, $X = \frac{x}{x^2 + y^2}$ gives (after cross multiplying) $x = X(x^2 + y^2)$, but $x^2 + y^2 = \frac1{X^2 + Y^2}$ (by taking reciprocals in the equation right above "Edit"), therefore:
$$x = X \times \frac{1}{X^2 + Y^2} = \frac{X}{X^2 + Y^2}$$
Similarly, we get:
$$y = \frac{Y}{X^2 + Y^2}$$
