In calculus it is usual to work with smooth functions in terms of `variables', I mean, something as $f(x,y,z)=x^2+\sin(z)+e^y$, in terms of `x', `y' and `z'. However working on surfaces, sometimes (or many), we prefer do not use `variables', specially when we need to compute the derivative of the function. The next example clarifies this issue.
If $S$ is a surface, define $$f:S\rightarrow{\mathbb R},\ f(p)=|p|^2=\langle p,p\rangle.$$ Here we use `p' instead of the variables. This function measures the square of the distance of the point $p$ to the origin of ${\mathbb R}^3$. We observe that if $p=(x,y,z)$, then $f(x,y,z)=x^2+y^2+z^2$, which is a known differentiable function in ${\mathbb R}^3$, but now $f$ is defined on a surface. If we want to prove that $f$ is differentiable on $S$, first we consider $F:{\mathbb R}^3\rightarrow{\mathbb R}$ the function $F(p)=\langle p,p\rangle$. Since $p\mapsto p$ is the identity, which is differentiable, then $F$ is noting the scalar product of a differentiable map by itself. Then $F$ is differentiable. Finally, $f=F_{|S}$, that is, the restriction on $S$ of a differentiable map of ${\mathbb R}^3$. This proves definitively that $f$ is differentiable.
Other example is the height function. Let $a\in {\mathbb R}^3$ be a unit vector and define $$f:S\rightarrow{\mathbb R},\ f(p)= \langle p,a\rangle.$$ This function measures the square of the distance of the point $p$ to the vector plane $\Pi$ orthogonal to $a$. For this reason, it is named height function. If we write in coordinates and $p=(x,y,z)$, we have $f(x,y,z)=a_1 x+a_2 y+a_3 z$, where $a=(a_1,a_2,a_3)$. If we want to prove that $f$ is differentiable on $S$ without the use of `x's', define $F:{\mathbb R}^3\rightarrow{\mathbb R}$ the function $F(p)=\langle p,a\rangle$. Since $p\mapsto p$ and $p\mapsto a$ are differentiable maps, then $F$ is the scalar product of two differentiable vector maps, so $F$ is differentiable. Finally, $f=F_{|S}$, proving that $f$ is differentiable.
