Show that $\text{grad }f=\frac{f_uG-f_vF}{EG-F^2}\textbf{x}_u+\frac{f_vE-f_uF}{EG-F^2}\textbf{x}_v$ 
The gradient of a differentiable function $f:S\to R$ is a differentiable map grad $f:S\to R^3$ which assigns to each point $p\in S$ a vector grad $f(p)\in T_p(S)\subset R^3$ such that
$$\langle \operatorname{grad}{f(p)}, v\rangle_p=df_p(v) \  \text{ for all } v\in T_p(S).$$

Show that

If $E,F,G$ are the coefficients of the first fundamental form in a parametrisation $\mathbf{x}:U\subset R^2\to S$, then grad $f$ on $\mathbf{x}(U)$ is given by
$$\text{grad }f=\frac{f_uG-f_vF}{EG-F^2}\mathbf{x}_u+\frac{f_vE-f_uF}{EG-F^2}\mathbf{x}_v.$$


In particular, if $S=R^2$ with coordinates $x,y$,
$$\operatorname{grad}{f}=f_xe_1+f_ye_2,$$
where $\{e_1,e_2\}$ is the canonical basis of $R_2$ (thus, the definition agrees with the usual definition of gradient in the plane).

The tools that I can think of right now are:

*

*The definition of $df_p(v)$, but the definition needs another map $\alpha:(-\epsilon,\epsilon)\to U$, and I think it is not necessary to introduce $\alpha$ in deriving $\operatorname{grad}{f}=\frac{f_uG-f_vF}{EG-F^2}\mathbf{x}_u+\frac{f_vE-f_uF}{EG-F^2}\mathbf{x}_v$.

*I also don't think we need to use the Jacobian in this case.

*Another doubt that I have is the expression that we are looking for $\operatorname{grad}{f}$ is inside the fundamental form $\langle \operatorname{grad}{f(p)}, v\rangle_p$ which we still need to find.

Can somebody give some clue on how to proceed?
 A: You only need to check that 
$$\langle \operatorname{grad}{f(p)}, v\rangle_p=df_p(v) $$
is satisfied for $v = \mathbf{x}_u$ and $\mathbf{x}_v$, as these two vectors spans $T_pS$. Note that $df_p(\mathbf{x}_u) = f_u$ (Do you know why?). On the other hand, 
$$\begin{split}
\langle \text{grad}(f), \mathbf{x}_u\rangle &=\left\langle \frac{f_uG-f_vF}{EG-F^2}\mathbf{x}_u+\frac{f_vE-f_uF}{EG-F^2}\mathbf{x}_v, \mathbf{x}_u \right\rangle \\
&= \frac{f_uG-f_vF}{EG-F^2} E + \frac{f_vE-f_uF}{EG-F^2} F \\
&=\frac{1}{EG-F^2} \left(f_u EG - f_v FE + f_v EF - f_u F^2 \right)\\
&= f_u\\
&=df_p(\mathbf{x}_u). 
\end{split}$$
The checking for $\mathbf{x}_v$ is similar. 
A: Since $\text{grad}f(p)\in T_pS$ and $\{\mathbf{x}_u,\mathbf{x}_v\}$ is a basis of $T_pS$, then exists $\alpha,\beta\in\Bbb R$ such that $\text{grad}f(p)=\alpha\cdot\mathbf{x}_u+\beta\cdot\mathbf{x}_v$. From this we have
\begin{align}f_u=df_p(\mathbf{x}_u)=\langle\text{grad}f(p), \mathbf{x}_u\rangle=\alpha\cdot\langle\mathbf{x}_u,\mathbf{x}_u\rangle+\beta\cdot\langle\mathbf{x}_v,\mathbf{x}_u\rangle=\alpha\cdot E+\beta\cdot F\\
f_v=df_p(\mathbf{x}_v)=\langle\text{grad}f(p), \mathbf{x}_v\rangle=\alpha\cdot\langle\mathbf{x}_u,\mathbf{x}_v\rangle+\beta\cdot\langle\mathbf{x}_v,\mathbf{x}_v\rangle=\alpha\cdot F+\beta\cdot G\end{align} 
So solving the linear system $\begin{bmatrix}E&F\\F&G\end{bmatrix}\begin{bmatrix}\alpha\\\beta\end{bmatrix}=\begin{bmatrix}f_u\\f_v\end{bmatrix}$ ends our problem. 
But since $\begin{bmatrix}E&F\\F&G\end{bmatrix}^{-1}=\dfrac{1}{EG-F^2}\begin{bmatrix}G&-F\\-F&E\end{bmatrix}$, follows that 
$$\begin{bmatrix}\alpha\\\beta\end{bmatrix}=\frac{1}{EG-F^2}\begin{bmatrix}G&-F\\-F&E\end{bmatrix}\begin{bmatrix}f_u\\f_v\end{bmatrix}=\begin{bmatrix}\dfrac{Gf_u-Ff_v}{EG-F^2}\\\dfrac{Ef_v-Ff_u}{EG-F^2}\end{bmatrix}.$$
That means $$\text{grad }f=\frac{f_uG-f_vF}{EG-F^2}\mathbf{x}_u+\frac{f_vE-f_uF}{EG-F^2}\mathbf{x}_v.$$
