Computing fundamental forms of implicit surface For an implicit surface $F(x,y,z)=0$ we can take $z=f(x,y)$ and use IFT to calculate its fundamental forms. But they are expressed by $dx^2,dxdy,dy^2$.
I wonder if there are other expressions using $dx^2,dy^2,dz^2,dxdy,dydz,dzdx$, just more symmetric.
Also I want to know how to differentiate entirely without taking coordinate parameters $x,y,z$ to get a direct expression by $dF$ with something.
Any help will be appreciated.
 A: I do not know the most efficient answer to the question that you are asking (in fact, the question is a little unclear), but the following addresses the question in the title.
Away from singular points, the gradient of $F$,  $\nabla F = F_{x}\frac{\partial}{\partial x} + F_{y} \frac{\partial}{\partial y} + F_{z}\frac{\partial}{\partial z}$, is perpendicular to the surface.
Further, at a point $P(a, b, c)$ on the surface $F(x, y, z) =0$, the equation of the tangent plane to the surface is given by
$$
F_{x}(P)(x - a) + F_{y}(P)(y - b) + F_{z}(P)(z - c)= 0.
$$
It is thus relatively easy to come up with an orthogonal basis for the tangent space to $\mathbb{R}^{3}$ at such points on the surface, but since you only know that one of the components of $\nabla F$ is non-zero, you need to work in cases.  For simplicity, work on an open set $U$ where the tangent plane is not horizontal so that you can assume that one of $F_{x}$ or $F_{y}$ is not-zero.
Then you can take the vector fields 
\begin{align*}
\vec{e}_{1} &= -F_{y}\frac{\partial}{\partial x}  + F_{x}\frac{\partial}
{\partial y}\\
\vec{e}_{2} &= -F_{x}F_{z}\frac{\partial}{\partial x}  - F_{y}F_{z}\frac{\partial}{\partial y} + (F_{x}^2 + F_{y}^2)\frac{\partial}{\partial z}\\
\vec{e}_{3} &= F_{x}\frac{\partial}{\partial x} + F_{y} \frac{\partial}{\partial y} + F_{z}\frac{\partial}{\partial z}
\end{align*}
to be a frame for the tangent space to $\mathbb{R}^{3}$ along the surface, with  $\vec{e}_{1}$ and $\vec{e}_{2}$ tangent to the surface and $\vec{e}_{3}$ perpendicular to the surface.
The one-forms dual to the frame above are thus
\begin{align*}
\omega^{1} &= -\frac{F_{y}}{A} dx + \frac{F_{x}}{A}dy\\
\omega^{2} &= -\frac{F_{x}F_{z}}{AB} dx - \frac{F_{y}F_{z}}{AB}dy + \frac{1}{B}dz\\
\omega^{3} &= \frac{F_{x}}{B}dx + \frac{F_{y}}{B} dy +\frac{F_{z}}{B} dz,\\
\end{align*}
where $A = F_{x}^2 + F_{y}^2$ and $B = F_{x}^2 + F_{y}^2 + F_{z}^2$,
and the standard inner product on $\mathbb{R}^3$ is then expressed as
$$
\mathbf{g} = A \omega^{1}\otimes \omega^{1} + AB \omega^{2}\otimes \omega^{2} + B \omega^{3}\otimes \omega^{3}.
$$
Restricted to the surface $S$ in question, one has
$$
\mathbf{g}\vert_{S} = A \omega^{1}\otimes \omega^{1} + AB \omega^{2}\otimes \omega^{2}.
$$
Substituting in the expressions for $\omega^{1}$ and $\omega^{2}$ one is then able to obtain an expression involving $dx^2, dxdz,$ etc.
Finally, normalizing the gradient vector field $\nabla F =  F_{x}\frac{\partial}{\partial x} + F_{y} \frac{\partial}{\partial y} + F_{z}\frac{\partial}{\partial z}$ one obtains the unit normal vector field along the surface expressed by
$$
\vec{n} = \frac{\nabla F}{\sqrt{B}} = \frac{F_{x}}{\sqrt{B}}\frac{\partial}{\partial x} + \frac{F_{y}}{\sqrt{B}} \frac{\partial}{\partial y} + \frac{F_{z}}{\sqrt{B}}\frac{\partial}{\partial z}.
$$
One should now be able to also express the second fundamental form $II$ relative to the chosen basis by computing
$$
II(\vec{e}_{1}, \vec{e}_{1}),\quad II(\vec{e}_{1}, \vec{e}_{2}),\quad \textrm{ and },\quad II(\vec{e}_{2}, \vec{e}_{2}),
$$
and obtaining 
$$
II = 
II(\vec{e}_{1}, \vec{e}_{1}) \omega^{1} \otimes \omega^{1} +  II(\vec{e}_{1}, \vec{e}_{2})\omega^{1} \otimes \omega^{2} + II(\vec{e}_{2}, \vec{e}_{1}) \omega^{2}\otimes \omega^{1} +  II(\vec{e}_{2}, \vec{e}_{2})\omega^{2}\otimes \omega^{2}.
$$
Substituting in the expressions for $\omega^{1}$ and $\omega^{2}$ one is then able to obtain an expression involving $dx^2, dxdz,$ etc.
Again, it is more than likely that there are more efficient ways to do this.
