I am working through an example to make sure I understand some theory about curvature of surfaces in $\mathbb R^d$. It looks like something is wrong. Could someone please advise?
The Theory: Suppose the manifold $M \subset \mathbb R^d$ has the form $\{x \in \mathbb R^d: F(x)=0\}$ for some some twice-differentiable function $F:\mathbb R^d \to \mathbb R$. The unit normal vector at $a \in M$ is $N(a) = \frac{\nabla F(a)}{ \|\nabla F(a)\|}$. By varying $a$ along a path $\gamma$ with $\gamma(0) = a$ we can define the derivative $D_{\gamma'(0)}N(a)$ of $N(x)$ in the $\gamma'(0)$ direction at $a$. It turns out the derivative is always contained in the tangent plane to $M$ at $a$. Hence we have a linear operator $\gamma'(0) \mapsto D_{\gamma'(0)}N(a)$ from the tangent space to itself.
To compute that operator use the chain rule on $N(x) = \frac{\nabla F(x)}{ \|\nabla F(x)\|}$ to get $\frac{\nabla ^2 F(x)}{ \|\nabla F(x)\|}$ plus some extra term that vanishes on the tangent space. It follows the derivative of $N(x)$ in the $\gamma'(0)$ direction at $a$ is nothing more than$ \frac{\nabla ^2 F(a)}{ \|\nabla F(a)\|}\gamma'(0)$.
The Example: Let $d=3$ and $F(x,y,z) = z-xy$. The normal direction at the point $a=(a,b,c)$ of the surface is $\nabla F(a) = (-b-a,1)$. The tangent space consists of all vectors $(x,y,z)$ perpendicular to the normal. Hence $0=(-b,-a,1) \cdot (x,y,z) = z - bx -ay \iff z=ay+bx$ and tangent space is $(x,y,ay+bx)$ for all $x,y \in \mathbb R$. The Hessian is $$\nabla ^2 F(a) =\begin{pmatrix} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}.$$
By the above the Hessian should preserve the tangent space and so $\nabla ^2 F(a)(z,y,ay+bx)$ is also perpendicular to $\nabla F(a)$. Let's check that: Multiplying by the matrix we get $\nabla ^2 F(a)(x,y,ay+bx) = (-y,-x,0)$. But this does not seem to be in the tangent space, as the product with $\nabla F(a)$ is $(-b-a,1)\cdot (-y,-x,0) = ax+by$ which is not obviously zero.
Where have I gone wrong?