Let $f:\mathbb{R}^3 \to \mathbb{R}^3$ be $$f(x,y,z) = \left(\frac{ix^2 + hy^2}{2}, \frac{jy^2+kz^2}{2}, \frac{mx^2 + nz^2}{2} \right).$$
My question is what restrictions are necessary on $i,h,j,k,m,n$ for there to be a neighborhood around a point $(x,y,z)$ that maps 1:1 onto an open set of the range space? Is the global mapping ever 1:1 and onto?
In understanding this function and how to apply the inverse function theorem, I'm not sure how to deal with the coefficients here. Is there a good textbook/resource that covers examples similar to this?
Given: $$\begin{gathered} f:{\mathbb{R}^3} \to {\mathbb{R}^3} \hfill \\ f({\text{x}},{\text{y}},{\text{z}}) = \frac{1}{2}\left( {\begin{array}{*{20}{c}} {\alpha \cdot {x^2} + \beta \cdot {y^2}} \\ {\gamma \cdot {y^2} + \delta \cdot {z^2}} \\ {\varepsilon \cdot {x^2} + \varphi \cdot {z^2}} \end{array}} \right) \hfill \\ \end{gathered} $$ Then: $$Df(x) = \left( {\begin{array}{*{20}{c}} {\alpha \cdot x}&{\beta \cdot y}&0 \\ 0&{\gamma \cdot y}&{\delta \cdot z} \\ {\varepsilon \cdot x}&0&{\varphi \cdot z} \end{array}} \right)$$ and $$\det (Df(x)) = (\alpha \gamma \varphi + \beta \delta \varepsilon ) \cdot xyz$$ So $\det (Df(x)) \ne 0$ if $\alpha \gamma \varphi + \beta \delta \varepsilon \ne 0$ and $xyz \ne 0$. In this case everything is regular.