Let $F$ be a homogeneous form with coefficients in $\mathbb{R}$. Suppose it defines a smooth projective variety, in other words at every point other than the origin at least one of the first partial derivatives is non-zero.


I am looking for $F$ which satisfies the following condition: There is a point in $P \in \mathbb{R}^n$ such that when we apply implicit function theorem and obtain a function $G$, all the second partial derivatives of $G$ is $0$ at $P$.


I have been trying to find such $F$ but I have not found one yet. If anybody has any leads I would appreciate it!


Clarification: Let $P= (p\_1, ..., p\_n)$. Since $\nabla F(P) \neq 0$ without loss of generality suppose $\partial F/\partial x\_n (P) \neq 0$. Then the implicit function theorem tells us that there exists some function $G$ satisfying $F(x\_1, .., x\_{n-1}, G(x\_1, .., x\_{n-1})) = 0$
for $(x\_1, .., x\_{n-1})$ in some open neighbourhood of $(p\_1, .., p\_{n-1})$.
I would like all the second partial derivatives of $G$ at $(p\_1, .., p\_{n-1})$ to be $0$.



Question about the implicit function theorem. an example of a homogeneous form for which its implicit function satisfies certain conditions
Here is a simple example: Take $F = w^3 +3 w u^2 -v^3$ on $\mathbb{R}^3$ with coordinates $(u,v,w)$. At the point $p=(u,v,w)=(1,0,0)$, we have that $F=0$ can be solved for $w$ as a function of $(u,v)$. Meanwhile, via implicit differentiation,
$$
w\_u = \frac{-2uw}{u^2+w^2}\quad\text{and}\quad w\_v = \frac{v^2}{u^2+w^2}
$$
while
$$
w\_{uu} = \frac{-2w(w^2{-}u^2)(w^2+3u^2)}{(u^2+w^2)^3},\quad 
w\_{uv} = \frac{2v^2(w^2{-}u^2)}{(u^2+w^2)^3}, 
$$
and
$$
w\_{vv} = \frac{2v(w^4{+}2w^2u^2{-}wv^3{+}u^4)}{(u^2+w^2)^3}.
$$
Clearly, all of these partials vanish at $p = (u,v,w) = (1,0,0)$.


**Remark:** Of course, the point $[p]=[1,0,0]$ in $\mathbb{RP}^2$ is a flex of the smooth cubic curve defined by $F=0$. Meanwhile, there are smooth quartic projective curves that have no (real) flexes, so they would not have any points of the kind you are seeking.


