Multivariable Application of inverse function theorem Consider the transformation $x=G(u)$  where $x=(u_{1},u_{2},u_{3})$ given by
$x_{1} = u_{1} +(u_{3})^{2}$
$x_{2} = u_{3} -(u_{1})^{2}$
$x_{3} = u_{1} +u_{2} + u_{3}$
1) compute the derivative of this transformation.
i am having a bit of trouble understanding what this all means
i know that i am supposed to be Computing Df from the other example in my textbook but the way $u_{1}$ and $x_{1}$ are defined has me completely baffled.
2) Use the inverse function theorem to show that the transformation is locally invertable if $u_{1}u_{2} > 0$ 
Could someone perhaps layout the inverse function theorem in a more palatable way to this question? my textbook gives only a proof of the theorem and im not sure how to apply it.

I suspect that you have an error in the definition of $G$ or the invertibility condition.
Let $G_1(u) = u_{1} +(u_{3})^{2}$,
 $G_{2}(u) = u_{3} -(u_{1})^{2}$,
$G_{3}(u) = u_{1} +u_{2} + u_{3}$.
Compute $DG(u) = \begin{bmatrix}
\frac{\partial G_1(u)}{\partial u_1} & \frac{\partial G_1(u)}{\partial u_2} & \frac{\partial G_1(u)}{\partial u_3} \\
\frac{\partial G_2(u)}{\partial u_1} & \frac{\partial G_2(u)}{\partial u_2} & \frac{\partial G_2(u)}{\partial u_3} \\
\frac{\partial G_3(u)}{\partial u_1} & \frac{\partial G_3(u)}{\partial u_2} & \frac{\partial G_3(u)}{\partial u_3} 
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 2 u_3 \\
-2 u_1 & 0 & 1 \\
1 & 1 & 1 
\end{bmatrix}
$.
Computing the determinant gives $\det( DG(u)) = -1 - 4 u_1 u_3$, hence $DG(u)$ is invertible iff $u_1 u_3 \neq - \frac{1}{4}$.
