Tangent Plane to a Surface

To find the tangent plane to a surface
\[F(x,y, z)=C\]
  at a point
\[(x_0 ,y_0 , z_0 )\]
, find the total derivative.
\[dF=\frac{\partial F}{\partial x}dx +\frac{\partial F}{\partial y}dy +\frac{\partial F}{\partial z}dz =0\]

Now put
\[dx=x-x_0 ,dy=y-y_0 , dz=z-z_0\]
to give
\[dF = \frac{\partial F}{\partial x}(x-x_0) +\frac{\partial F}{\partial y}(y-y_0) +\frac{\partial F}{\partial z}(z-z_0) =0\]

Example:
\[F(x,y, z)=x^2 y-z=-1\]
  at the point
\[(1,2,3)\]

\[ \begin{equation} \begin{aligned} dF_{(1,2,3)} &=(2xy)_{(1,,2,3)}dx +(x^2)_{(1,,2,3)} dy +(-1)_{(1,2,3)} \\ &=4dx+dy-dz \\ &=4(x-1)+(y-2)-(z-3) \\&=4x+y-z+3=0 \end{aligned} \end{equation}\]

Hence
\[4x+y-z=-3\]