second fundamental form of a surface given by regular values Consider the differentiable function $f:\mathbb R^3 \to \mathbb R$. Consider the regular surface given by $ f(x,y,z)=a$ where $a$ is a regular value of $f$. Prove that the second fundamental form of the surface is given by:
$$
\text{II} \left( v \right) = \frac{{Hess\left( f \right)\left( {v,v} \right)}}
{{\left| {\left| {\nabla f} \right|} \right|}}
$$
I only proved that the normal vector is given by $$
N  = \frac{{ \pm \nabla f}}
{{\left| {\left| {\nabla f} \right|} \right|}}
$$
This is clearly if we consider a curve $( x(t),y(t),z(t))$ lying on the surface. We have that $ f( x(t),y(t),z(t)) = a$ , that implies that $$
\left( {\nabla f} \right) \cdot \left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right) = f_x x' + f_y y' + f_z z' = 0
$$
So one way to do it , it's computing the Jacobian matrix of N$ but I think that that there is a more "smart" way to do it
 A: Well, it should be a fact you recall from multivariable calculus that the gradient gives the normal vector to a level surface. So you want to compute (all at a point $p$) $\text{II}(\vec v) = -dN(\vec v)\cdot\vec v$. Choose a path $\vec x(t)$ with $\vec x(0) = p$ and $\vec x{}'(0) = \vec v$. Then
$$\text{II}(\vec v) = -(N\circ \vec x)'(0)\cdot \vec x{}'(0) \,.$$
By the chain rule and the product rule the derivative of $N\circ\vec x=\dfrac{\nabla f}{\|\nabla f\|}\circ\vec x$ at $t=0$ is 
$$\dfrac1{\|\nabla f\|} \text{Hess}(f)\vec x{}'(0) + (\dots)\nabla f\,,\tag{$\star$} 
$$
where $f$ and its derivatives are evaluated at $\vec x(0) = p$.
Recalling that $\vec x{}'(0) = \vec v$, we have (up to a sign)
$$\text{II}(\vec v) = \dfrac1{\|\nabla f\|} \text{Hess}(f)(\vec v,\vec v) + (\dots)\nabla f\cdot\vec v= \dfrac1{\|\nabla f\|} \text{Hess}(f)(\vec v,\vec v)\,,$$
since $\nabla f\cdot\vec v = 0$. (Note that in ($\star$) I'm writing the product of the Hessian matrix with the vector $\vec v$, and then in your original equation $\text{Hess}(f)(\vec v,\vec v)$ is the dot product of that vector with $\vec v$.)
