I am trying to derive a formula for the tangent plane to a surface at $(x_0,y_0,z_0)$.
I started with $F(x,y,z)=0$ for $(x,y,z)$ near and at $(x_0,y_0,z_0)$. It can be seen that any curve in the surface that passes through and has a tangent at $(x_0,y_0,z_0)$ has a position vector $x(t)i+y(t)j+z(t)k$ for $t$ near $0$, and $t=0$ gives the position vector of $(x_0,y_0,z_0)$; also, any vector that is orthogonal to $x'(0)i+y'(0)j+z'(0)k$ is a normal vector of the tangent plane.
It then seems like we can differentiate $F(x(t),y(t),z(t))=0$ with respect to $t$ at $t=0$ using the chain rule. But I do not think that is okay to do because the chain rule needs $F$ being differentiable at $(x_0,y_0,z_0)$ which is not easy to see.