Let $K = \{ (x,y,z) \mathbb{R}^3 | (\frac xa)^2 +(\frac yb)^2 +(\frac zc)^2 =1\}$ oriented by outward-pointing unit normal field. Consider $f(x,y,z) = ( \frac {ay}b, \frac{bz}c, \frac{cx}a)$, $f: K \mapsto K$. I am asked to show whether $f$ is orientation preserving or orientation reserving and for $a>b>c>0$, want to find maximal $|| df_p||$ (area distortion). Area distortion is defined as $$\frac{|df_p(v_1) \times df_p(v_2)|}{|v_1 \times v_2|},$$ where $v_1,v_2$ is a basis of $K$.
My attempt: $$df_p = \begin{pmatrix} 0 & 0& \frac ac \\ \frac ba & 0 & 0 \\ 0 & \frac cb & 0 \end{pmatrix}$$, $\det(df_p)=1$ for every $p$. So it is orientation preserving, and since sign of determinant does not change with respect to basis we choose, it is orientation preserving. Furthermore since $df_p(v_1) \times df_p(v_2) = \det(df_p) ( v_1 \times v_2)$, the area distortion does not change so every $p \in K$ is maximal and minimal.
But I feel like there is something I am missing because I did not use outward-pointing unit normal field, and since area distortion is the same everywhere, there is no maximal. Is anything wrong with what I've done?