\[F(x,y, z)=C\]
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
\[4x+y-z=-3\]
The equation
\[4dx+dy-dz=0 \]
can be written \[(4 \mathbf{i}+1 \mathbf{j}+(-1)\mathbf{k}) \cdot (dx\mathbf{i}+ dy \mathbf{j}+ dz \mathbf{k})=0 \]
.Hence
\[4 \mathbf{i}+\mathbf{j}-\mathbf{k}\]
is normal to the surface at the the point \[(1,2,3)\]
.The normal line is then
\[1 \mathbf{i}+2 \mathbf{j}+3 \mathbf{k}+t(4 \mathbf{i}+\mathbf{j}-\mathbf{k}) \]