Let $G:{\mathbb{R}^3} \to {\mathbb{R}^3}$ be defined by $G\left( {\rho ,\theta ,\phi } \right) = \left( {\rho \cos \theta \sin \phi ,\rho \sin \theta \sin \phi ,\rho \cos \phi } \right)$.
I've found that ${G_{\rho \theta \phi }}\left( {\rho ,\theta ,\phi } \right) = {G_{\theta \phi \rho }}\left( {\rho ,\theta ,\phi } \right) = \left( { - \sin \theta \cos \phi ,cos\theta \cos \phi ,0} \right)$.
Since ${G_{\rho \theta \phi }}\left( {\rho ,\theta ,\phi } \right) = {G_{\theta \phi \rho }}\left( {\rho ,\theta ,\phi } \right)$, can someone please tell me what I can conclude about the map $G$?
Notice that you don’t need to go to third derivatives: even just mixed second derivatives end up being the same. You may conclude that the matrix consisting of all second partial derivatives of $G$ (its Hessian) is symmetric, defining a smooth (possibly degenerate and not necessarily positive definite) metric form on the domain of $G$. The spectral theorem implies that, at any point in the domain, there exist orthogonal principal axes for this form and it is diagonalizable (with real eigenvalues). The sign of these eigenvalues, which correspond to pure second directional derivatives in the directions determined by the principal axes at that point, will tell you about the local convexity of $G$. Notice that, by Sylvester’s law of inertia, the signature of the Hessian at a point will not depend on the particular matrix used to render it diagonal.