I'm trying to prove that in $RP^2$, given a homogeneous polynomial $F$ of degree $k$, the hypersurface $Z(F)$ of the zeros of $F$ is smooth if $\frac{\partial F}{\partial x_0}$, $\frac{\partial F}{\partial x_1}$ , $\frac{\partial F}{\partial x_2}$ are not all simultaneously $0$ on $Z(F)$.
What I tried to do is to show that $Z(F)$ is a submanifold of $RP^2$: given a point $p=[x_0 x_1 x_2] \in RP^2$ w.l.o.g. assume $x_0 \neq 0$ so take the neigboorhood $U_0$ of points with $x_0 \neq 0$. In $U_0$, $Z(F)$ is given by the zeros of $f(x,y)$ which is defined as $f(x,y):=F(1,x_1/x_0,x_2/x_0)$. My idea is to prove that the rank of $f$ is 1 on $Z(F)$, and then in the new coordinates of $U_0$ (assuming $\frac{\partial f}{\partial x} \neq 0 $), $(f,y)$ the set $Z(F)$ is given by $f=0$.
So all the problem is reduced to prove that either $\frac{\partial f}{\partial x} \neq 0 $ or $\frac{\partial f}{\partial y} \neq 0 $ in $Z(F)$.
In a point $(\alpha,\beta)=[1,\alpha,\beta]$ of $Z(F)$,
$\frac{\partial f}{\partial x} (\alpha,\beta)= \frac{\partial F}{\partial x_1} ([1,\alpha,\beta]) \frac{\partial x_1}{\partial x}(1,\alpha) = \frac{\partial F}{\partial x_1} ([1,\alpha,\beta]) $. Similarly,
$\frac{\partial f}{\partial y} (\alpha,\beta)= \frac{\partial F}{\partial x_2} ([1,\alpha,\beta]) $,
and since $\frac{\partial F}{\partial x_0}$, $\frac{\partial F}{\partial x_1}$ , $\frac{\partial F}{\partial x_2}$ are not all simultaneously $0$ on $Z(F)$, $f$ has rank 1 on $Z(F)$.
Is this argument correct? I don't understand the meaning of $\frac{\partial F}{\partial x_i} $ since $F$ is not a function on $RP^2$.
Generally, to show that $N$ is an embedded submanifold of $M$ it suffices to exhibit an open cover $U_j$ of $M$ such that $N\cap U_j$ is an embedded submanifold of $M\cap U_j$ for every $j$.
A natural choice of such a cover for $\mathbb RP^2$ is $U_j=\{[x_0:x_1:x_2] : x_j\ne 0\}$, $j=0,1,2$. On $U_0$ we have the diffeomorphism $\Phi:U_0\to \mathbb R^2$ given by $[x_0:x_1:x_2]\mapsto (x_1/x_0,x_2/x_0)$. It maps the zero set of $F$ to the zero set of the function $f(x,y)=F(1,x,y)$. Since the gradient of $f$ does not vanish (as you've shown), the zero set is a submanifold.
(CW posted to fill this box)