Let $S$ be the (level) surface $z\=z\left(x\,y\right)$ defined implicitly by $f\left(x\,y\,z\right)\=c$.

Let P:$\left(\mathrm{\α}\,\mathrm{\β}\,\mathrm{\γ}\right)$ be a point on on $S$ so that $f\left(\mathrm{\α}\,\mathrm{\β}\,\mathrm{\γ}\right)\equiv c$.

Let ${\mathbf{C}}_1\=\left[\begin{array}{c}\mathrm{\α}\\ y\\ z\(\mathrm{\α}\,y\)\end{array}\right]$ describe the coordinate curve in $S$ that projects onto the grid line $x\=\mathrm{\α}$.

Let ${\mathbf{C}}_2\=\left[\begin{array}{c}x\\ \mathrm{\β}\\ z\(x\,\mathrm{\β}\)\end{array}\right]$ describe the coordinate curve in $S$ that projects onto the grid line $y\=\mathrm{\β}$.

Then ${\mathbf{T}}_1\=\left[\begin{array}{c}0\\ 1\\ z_y\end{array}\right]$ and ${\mathbf{T}}_2\=\left[\begin{array}{c}1\\ 0\\ z_x\end{array}\right]$ are tangent respectively to ${\mathbf{C}}_1$ and ${\mathbf{C}}_2$.

The gradient $\nabla f$ will be orthogonal to $S$ if it is orthogonal to both ${\mathbf{T}}_1$ and ${\mathbf{T}}_2$.

Implicitly differentiate $f\left(x\,y\,z\left(x\,y\right)\right)\equiv c$ to get $f_x\+f_z z_x\=0$ and $f_y\+f_z z_y\=0$, from which $z_x\=-f_x\/f_z$ and $z_y\=-f_y\/f_z$ then follow.

Thus, ${\mathbf{T}}_1$ and ${\mathbf{T}}_2$ become respectively $\left[\begin{array}{c}0\\ 1\\ -f_y\/f_z\end{array}\right]$ and $\left[\begin{array}{c}1\\ 0\\ -f_x\/f_z\end{array}\right]$, so that

Hence, $\nabla f$ is orthogonal to two independent tangent vectors on $S$, so it is orthogonal to $S$ itself.