This is a problem from Apostol, Calculus, Volume II (p. $302$, Chapter $9.8$).
The three equations $F(u,v) = 0$, $u = xy$, and $v = \sqrt{x^2 + z^2}$ define a surface in $xyz$-space. Find a normal vector to this surface at the point $x = y = 1, z = \sqrt{3}$ if it is known that $D_1F(1,2) = 1$ and $D_2 F(1,2) = 2$.
I think we would like to calculate $\partial F/\partial x$, $\partial F/\partial y$, $\partial F/\partial z$, then the normal is given by the function with these as its component functions and we can evaluate at $x=y=1$, $z = \sqrt{3}$. First, we calculate
\begin{align*} \frac{\partial F}{\partial x} &= \frac{\partial F}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial F}{\partial v}\frac{\partial v}{\partial x} &&= y \frac{\partial F}{\partial u} + \frac{x}{\sqrt{x^2+z^2}}\frac{\partial F}{\partial v}\\ \frac{\partial F}{\partial y} &= \frac{\partial F}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial F}{\partial v}\frac{\partial v}{\partial y} &&= x \frac{\partial F}{\partial u}\\ \frac{\partial F}{\partial z} &= \frac{\partial F}{\partial u}\frac{\partial u}{\partial z} + \frac{\partial F}{\partial v}\frac{\partial v}{\partial z} &&= \frac{z}{\sqrt{x^2+z^2}}\frac{\partial F}{\partial v} \end{align*}
From here, I do not know how to proceed. The answer given in the book is
$$ 2\mathbf{i} + \mathbf{j} + \sqrt{3} \mathbf{k}. $$
Define the function $G:\mathbb{R}^3\rightarrow\mathbb{R}^2$ to be $G(x,y,z)=(u,v)=(u(x,y,z),v(x,y,z))$ and the function $H=F\circ G:\mathbb{R}^3\rightarrow\mathbb{R}$. Since your surface is defined by $H=0$, the normal to this surface at any given point $(x_0,y_0,z_0)$ is the vector $\nabla H(x_0,y_0,z_0)$. For example $\partial_x H$ is given by \begin{equation} \partial_x H(x,y,z)=\partial_u F(u,v)\partial_xu(x,y,z)+\partial_v F(u,v)\partial_xv(x,y,z) \end{equation} as you calculated. You have: \begin{equation} \partial_xu=y\qquad\partial_yu=x\qquad\partial_zu=0 \end{equation} and \begin{equation} \partial_xv=\frac{x}{\sqrt{x^2+z^2}}\qquad\partial_yv=0\qquad\partial_zv=\frac{z}{\sqrt{x^2+z^2}} \end{equation} At the point $(1,1,\sqrt{3})$ you get $u=1,v=2$; $\partial_xF(1,2)=1$ and $\partial_yF(1,2)=2$: \begin{equation} \partial_x H(1,1,\sqrt{3})=\partial_u F(1,2)\partial_xu(1,1,\sqrt{3})+\partial_v F(1,2)\partial_xv(1,1,\sqrt{3}) \end{equation} which yelds \begin{equation} \partial_x H(1,1,\sqrt{3})=1\cdot\partial_xu(1,1,\sqrt{3})+2\cdot\partial_xv(1,1,\sqrt{3})\\=1\cdot 1 + 2\cdot\frac{1}{2}=2 \end{equation} So the first component of your vector is $2$. The others are obtained in the same way.