Smoothness of projective hypersurface I'm trying to understand the question and answer here, but I don't quite follow what they're doing, so here is my take on it. The problem is to show that in $\mathbb R P^2$, 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 zero on $Z(F)$.
Letting $U_{0} =
\{[x_{0},x_{1},x_{2}]\in \mathbb R P^2 \mid x_0\neq 0\}$, the standard coordinates on $U_0$, which is homeomorphic to $\mathbb R^2$, are $x = x_1/x_0$, $y = x_2/x_0$. Now in $U_0$,
$$
    F(x_0, x_1, x_2) = x_0^k F(1, x_1/x_0, x_2/x_0) = x_0^k F(1,x,y),
$$
so if we define $f(x,y) = F(1,x,y)$, $f$ and $F$ have the same zero set in $U_0$.
While $F$ is not a well-defined function on $\mathbb R P^2$, I think that $f$ is well-defined on $U_0$ since the choice of $x$ and $y$ is unique?
Now let $(\alpha,\beta)\in U_0\cap Z(f)$. We know that either $\frac{\partial F}{\partial x_0}(1,\alpha,\beta)$, $\frac{\partial F}{\partial x_1}(1,\alpha,\beta)$, or $\frac{\partial F}{\partial x_2}(1,\alpha,\beta)$ is nonzero. Apparently we have that
\begin{align*}
\frac{\partial f}{\partial x}(\alpha,\beta) & = \frac{\partial F}{\partial x_1}(1,\alpha,\beta) \\
\frac{\partial f}{\partial y}(\alpha,\beta) & = \frac{\partial F}{\partial x_2}(1,\alpha,\beta),
\end{align*}
but I'm not entirely seeing why; it seems like some kind of chain rule stuff? And then we want to conclude that either $\frac{\partial f}{\partial x}(\alpha,\beta)$ or $\frac{\partial f}{\partial y}(\alpha,\beta)$ is nonzero, but I don't see exactly how we could conclude that, because couldn't we have that $\frac{\partial F}{\partial x_1}(1,\alpha,\beta) = \frac{\partial F}{\partial x_2}(1,\alpha,\beta) = 0$?
I would appreciate some advice to clarify the above; in particular the places where I inserted question marks.
 A: There's really no chain rule going on here. If you start with a function $F(x_1,x_2)$ and you set $f(x) = F(x,3)$, then by the definition of partial derivative, $f'(x) = \dfrac{\partial F}{\partial x_1}(x,3)$. Similarly in this case.
The key answer to your question is Euler's Theorem on Homogeneous Functions. If $F(x) = F(x_0,x_1,x_2)$ is homogeneous of degree $k$, then (it is an exercise in the chain rule to show that)
$$\sum_{i=0}^2 x_i\frac{\partial F}{\partial x_i} = k F(x_0,x_1,x_2).$$
Thus, on the zero-set of $F$, we have $\sum_{i=0}^2 x_i\frac{\partial F}{\partial x_i} = 0$. This means that the gradient vector $\nabla F(x)$ is the normal vector to the tangent plane of $F=0$. Because $F=0$ is a cone, the tangent plane always contains the ruling of the cone, i.e., the line spanned by $x$. (To be official, if $\pi: \Bbb R^3-\{0\}\to\Bbb RP^2$ is the projection, the tangent line to the curve $F=0$ in $\Bbb RP^2$ is the image of that tangent plane under the derivative mapping $D\pi(x): T_x(\Bbb R^3-\{0\}) \to T_{\pi(x)}(\Bbb RP^2)$. This map collapses the ruling to $0$.)
Note, in particular, that at a point with $x_0\ne 0$, we cannot have $\dfrac{\partial F}{\partial x_0} \ne 0$, but $\dfrac{\partial F}{\partial x_1} = \dfrac{\partial F}{\partial x_2} = 0$. Think about this in terms of the geometry of the cone I alluded to earlier.
