$P=[x_0:y_0:z_0]$ is a singularity of $F=0$ iff $\frac{\partial F}{\partial x}(P)=\frac{\partial F}{\partial y}(P)=\frac{\partial F}{\partial z}(P)=0$

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$F(x,y,z)$ is a homogeneous polynomial of degree $3$.

Use the identity $d*F=x\frac{\partial F}{\partial x}+y\frac{\partial F}{\partial y}+z\frac{\partial F}{\partial z}$ to show that a point $P=[x_0:y_0:z_0]\in \mathbb P^2$ is a singular point of the plane curve $F(x,y,z)=0$ if and only if $\frac{\partial F}{\partial x}(x_0,y_0,z_0)=\frac{\partial F}{\partial y}(x_0,y_0,z_0)=\frac{\partial F}{\partial z}(x_0,y_0,z_0)=0$.

I know the definition of the singular point: If every line through P intersects F at least twice at P(i.e $I_P(\ell ,F)\ge2$), then P is a singular point.

I know the tangent plane through P is $\nabla F(P)\cdot(x,y,z)=0$

I don't know what the relation is between the identity and the definition.