Finding a normal vector to the surface $F(u,v)=0. u=xy, v = \sqrt {x^2+z^2}$ at the point $x=1,y=1, z=\sqrt 3$ The three equations $F(u,v)=0. u=xy, v = \sqrt {x^2+z^2}$ define a surface in $xyz$ space. Find a normal vector to this surface at the point $x=1,y=1, z=\sqrt 3$ if it is known that $D_1F(1,2)=1$ and $D_2F(1,2)=2$
Solution Attempt:
Since, $F(u,v)=0$ represents a level surface, hence, the gradient vector will be a normal vector to the surface at a given point.
$F(u,v)=0 \implies \dfrac {\partial F}{\partial u} \cdot  \dfrac {\partial u}{\partial x} +  \dfrac {\partial F}{\partial v} \cdot  \dfrac {\partial v}{\partial x} =0$
$\implies \dfrac {\partial F}{\partial u} \cdot  y +  \dfrac {\partial F}{\partial v} \cdot  \dfrac  {x} {\sqrt {x^2+z^2}} =0$
At the given point, this becomes equal to :
$ \dfrac {\partial F}{\partial u}  +  \dfrac {\partial F}{\partial v} \cdot  \dfrac  {1} {2} =0~~~~~........(1)$ 
Similarly: $\dfrac {\partial F}{\partial u} \cdot  \dfrac {\partial u}{\partial y} +  \dfrac {\partial F}{\partial v} \cdot  \dfrac {\partial v}{\partial y} =0$
$\implies \dfrac {\partial F}{\partial u} \cdot  x +  \dfrac {\partial F}{\partial v} \cdot  0 =0~~~~~......(2)$ 
From $(1),(2)$, we get that $ \dfrac {\partial F}{\partial u}= \dfrac {\partial F}{\partial v} = 0  $
This means, that the normal vector is $0$?
Could anyone please tell me where I could have possibly gone wrong in my solution attempt? 
Thank you for reading through and helping!
 A: If $\vec{n}$ is the normal vector at $P_0=(1,1,\sqrt3)$ then
$$\vec{n}=\frac{\nabla F}{|\nabla F|}$$
We just need to find $\nabla F$ at $P_0$
$$\nabla F =\dfrac {\partial F}{\partial x}\vec{i}+\dfrac {\partial F}{\partial y}\vec{j}+\dfrac {\partial F}{\partial z}\vec{k}$$
$$\nabla F =(\dfrac {\partial F}{\partial u}\dfrac {\partial u}{\partial x}+\dfrac {\partial F}{\partial v}\dfrac {\partial v}{\partial x})\vec{i}+(\dfrac {\partial F}{\partial u}\dfrac {\partial u}{\partial y}+\dfrac {\partial F}{\partial v}\dfrac {\partial v}{\partial y})\vec{j}+(\dfrac {\partial F}{\partial u}\dfrac {\partial u}{\partial z}+\dfrac {\partial F}{\partial v}\dfrac {\partial v}{\partial z})\vec{k}$$
$$\nabla F =(D_1F\cdot\dfrac {\partial u}{\partial x}+D_2F\cdot\dfrac {\partial v}{\partial x})\vec{i}+(D_1F\cdot\dfrac {\partial u}{\partial y}+D_2F\cdot\dfrac {\partial v}{\partial y})\vec{j}+(D_1F\cdot\dfrac {\partial u}{\partial z}+D_2F\cdot\dfrac {\partial v}{\partial z})\vec{k}$$
$$\nabla F =(1\cdot\dfrac {\partial u}{\partial x}+2\cdot\dfrac {\partial v}{\partial x})\vec{i}+(1\cdot\dfrac {\partial u}{\partial y}+2\cdot\dfrac {\partial v}{\partial y})\vec{j}+(1\cdot\dfrac {\partial u}{\partial z}+2\cdot\dfrac {\partial v}{\partial z})\vec{k}$$
$$\nabla F =(y+\frac{2x}{\sqrt{x^2+z^2}})\vec{i}+(x+0)\vec{j}+(0+\frac{2z}{\sqrt{x^2+z^2}})\vec{k}$$
at poit $P$:
$$\nabla F =2\vec{i}+1\vec{j}+\sqrt3\vec{k}$$
$$|\nabla F| =\sqrt{8}=2\sqrt{2}$$
and
$$\vec{n}=\frac{\nabla F}{|\nabla F|}=(\frac{\sqrt2}{2},\frac{\sqrt2}{4},\frac{\sqrt6}{4})$$
