Let $S$ be a regular surface and $f:S \to R$ be a smooth function on $S$. Define a vector field $\bigtriangledown f$ by $\bigtriangledown f(p).v = f_{*p}(v) \in \Bbb R \ \ \forall v \in T_pS$. Here $f_{*p}$ denotes the differential of $f$ or the pushforward map.
Let $E,F,G$ denotes the components of the first fundamental form in a coordinate chart $(\sigma ,U)$.
To show that $$\bigtriangledown f |_{\sigma(U)} = \frac{f_uG - f_vF}{EG-F^2} \sigma_u+ \frac{f_vE - f_uF}{EG-F^2} \sigma_v$$
where $f_u= \frac{\partial f \circ \sigma(u,v)}{\partial u}$ and $f_v = \frac{\partial f \circ \sigma(u,v)}{\partial v}$.
Require some hints t proceed with the problem.
Since $\{\sigma_u,\sigma_v\}$ is a basis of $TS_{\vert U}$, there exist smooth maps $a,b\colon U\rightarrow\mathbb{R}$ such that $\nabla f=a\sigma_u+b\sigma_v$. Plugging $v=\sigma_u,\sigma_v$ in the equation $\mathrm{d}f(v)=\langle\nabla f,v\rangle$ leads to the following system of equations: $$\left\{\begin{align}f_u&=aE+bF,\\f_v&=aF+bG,\end{align}\right.$$ recalling that $E=\|\sigma_u\|^2$, $F=\langle\sigma_u,\sigma_v\rangle$ and $G=\|\sigma_v\|^2$. Whence the result since $\begin{pmatrix}E & F\\F&G\end{pmatrix}$ is invertible.