A problem on projective curve. Let $P$ be a point on the projective plane curve $C=C_F$. Show that $P$ is singular on the affine plane curve $C_i$ for some $i$ iff 
$$F(P)=0=\Big(\frac{\partial F}{ \partial X}\Big)_P=\Big(\frac{\partial F}{ \partial Y}\Big)_P=\Big(\frac{\partial F}{ \partial Z}\Big)_P.$$
When $P$ is non-singular, show that the projective line
$$L: \Big(\frac{\partial F}{ \partial X}\Big)_PX+\Big(\frac{\partial F}{ \partial Y}\Big)_PY+\Big(\frac{\partial F}{ \partial Z}\Big)_PZ=0$$
has the property that $L\cap U_i$ is the tangent line at $P$ to the affine curve $C_i$ for all $i$ such that $P$ lies in $U_i$; where $U_0=\{(x:y:z)| x \neq 0\}$ and so on.
My try: $'\Rightarrow'$ Part
As, $P \in C_F$ we have $F(P)=0$. Now suppose $Z=c(say)$-co-ordinate of $P$ is non-zero then $$\Big(\frac{\partial F}{ \partial X}\Big)_P=\Big(\frac{\partial F}{ \partial Y}\Big)_P= {c^n}\Big(\frac{\partial F(X/Z,Y/Z,1)}{ \partial X}\Big)_P= {c^n}\Big(\frac{\partial F(X/Z,Y/Z,1)}{ \partial Y}\Big)_P=0$$
Now if $P=(a,b,c)$ where $b,c \neq 0$ then playing the same trick I can prove that $$\Big(\frac{\partial F}{ \partial Z}\Big)_P=0$$
Otherwise $P=(0,0,c)$ then $F(P)=0$ implies that there is a polynomial $F(0,0,Z)$ in $Z$ such that $c$ is a root of this and $F$ is homogeneous so it will have only one coefficient say $F(0,0,Z)_P=c Z^{deg F}=0 \Rightarrow c=0$ then $$\Big(\frac{\partial F}{ \partial Z}\Big)_P=0$$.
So one part is done.
$\Leftarrow$ is rather easy as the curve is singular at $P$ at all canonical affine plane.
I need help for the $2$nd part.
 A: Let $P=(a:b:c)$ be a nonsingular point on a curve $C=V(F)\subset\mathbb P^2$. Consider the affine cone over it: $\mathcal C=V(F)\subset\mathbb A^3$. The tangent plane to $\mathcal C$ at the point $p=(a,b,c)$ has the equation
$$\Big(\frac{\partial F}{ \partial X}\Big)_P(X-a)+\Big(\frac{\partial F}{ \partial Y}\Big)_P(Y-b)+\Big(\frac{\partial F}{ \partial Z}\Big)_P(Z-c)=0.$$
The Euler's theorem on homogeneous functions says that $\frac{\partial F}{ \partial X}X+\frac{\partial F}{ \partial Y}Y+\frac{\partial F}{ \partial Z}Z=\mathrm{deg}\,F\cdot F$, therefore the above equation is equal to
$$\Big(\frac{\partial F}{ \partial X}\Big)_PX+\Big(\frac{\partial F}{ \partial Y}\Big)_PY+\Big(\frac{\partial F}{ \partial Z}\Big)_PZ=0,$$
which defines a projective line $L$. 
Suppose that $a\neq 0$, so that $P$ lies on the affine plane curve $C_0$. Now $C_0$ is the intersection of $\mathcal C$ with the plane $H_0:X-1=0$, and $L\cap U_0$ is equal to $L\cap H_0$. Because homothety preserves tangency, we can instead consider intersections with the plane $H_{0,a}:X-a=0$: $L\cap H_0$ is tangent to $\mathcal C\cap H_0$ at the point $(1,\frac ba,\frac ca)$ if and only if $L\cap H_{0,a}$ is tangent to $\mathcal C\cap H_{0,a}$ at $P$. The latter follows from a simple statement that if we intersect a surface and its tangent plane with another plane through the point of contact we get a curve and its tangent line.
