Let $f: \mathbb{R}^{2} \to \mathbb{R}^{3}$ defined by $f(x,y) = (\sin x \cos y, \sin x \sin y, \cos x)$.
This question is divided into several items, one of them is
Show that $\parallel Df(x,y)\parallel_{2,2} = 1$ for any $(x,y) \in \mathbb{R}^{2}$.
Until now, it wasn't necessary to calculate explicitly the derivative of $f$ and I'd like to avoid unnecessary calculations, so I'm trying to prove it without this calculation. It's easy to see that $\parallel f(x,y) \parallel_{2,2} = 1$. Using this, is it possible to prove this item? Otherwise, is there any clever way to calculate this derivative?
Notation. $\parallel \cdot \parallel_{2,2}$ denotes the norm induced by the Euclidean norms of $\mathbb{R}^{2}$ and $\mathbb{R}^{3}$.
It would be indeed nice to have a completely "coordinate free" proof of the fact that the differential of $f$ has norm $1$ as a map from $\ell_2^2$ to $\ell_2^3$. But avoiding the calculation of the tangent vectors to the sphere at the point $f(x,y)$,misses out some essential information - that is - that the Euclidean norm of of the partial derivative $\frac{\partial f}{\partial x}$ has norm $1$ for every $x,y$, whereas the partial derivative $\frac{\partial f}{\partial y}$ has norm $|\sin x|$ for every $(x,y)$. The fact that the norm of $Df$ is at most $1$ now easily follows from the Cauchy-Schwartz inequality. That the norm is precisely $1$ follows by examining the image of $(1,0)$, say, under the differential.