Implicit Differentiation problem in Apostol Calculus vol.II 9.8.5 Exercise 9.8.5 from Apostol Calculus vol II:

I'm having problems understanding how to solve this apparently straightforward problem.
It asks for a normal vector to the surface defined by the 3 equations F(u,v). I know the normal vector of a surface is its gradient. But I'm having problems wrapping my head around this surface. I guess that the point exercise is to infer the gradient without and explicit definition of the surface. But how to do this exactly?
 A: *

*Any differentiable map induces a linear map on the tangent spaces i.e if  $f: U \to V$ and $p \in U, f(p) \in V$ then $Df(p): T_pU \to T_{f(p)}V$ defined by $[\gamma] \mapsto [f\circ \gamma]$. 

*The notation means, "for any point, you have a curve $\gamma$ going through it, let $[\gamma]$ be the velocity of $\gamma$ at $t_0$ where $\gamma(t_0) = p$. It follows that if $U,V$ are euclidean open sets then,
$$Df(p) = \begin{bmatrix} \frac{\partial f_i}{\partial x_j}(p) \end{bmatrix}_{i,j}$$


*

*Similarly we have $[f\circ \gamma]$ is the velocity vector of $f \circ \gamma$ at $p$. We use the equivalence class notation because any other differentiable function defined in a neighborhood of $p$ which agrees with $\gamma$, will give the same velocity vector. Here you have your surface $X = F^{-1}(\{0\})$ and so $F(X) = 0$ i.e, 


$$DF(\textbf{x}): T_{\textbf{x}}U \to T_{F(\textbf{x})}\mathbb{R}: DF(\textbf{x}) (\vec{v})= \langle \nabla f(\textbf{x}),\vec{v} \rangle$$
where the angle bracket denotes the standard dot product.


*

*Furthermore we have $\ker(DF(\textbf{x})) = \{\vec{v}: \langle\nabla f(\textbf{x}),\vec{v} \rangle = 0\}$. Now what fact from multivariate calculus can you use about the gradient vector of a differentiable real-valued function in relation to the tangent plane? 

*If you wish to use implicit differentiation, there's a reason why I wrote $F(X) = 0$.
