Suppose you have a vector $v$ in the tangent plane at a point $x$ on the surface. Draw a curve on the surface $\phi (t)$ going through $x$ at $t=0$ with tangent vector $v$ at $x$, i.e. $\phi <\mathrm{quote}/>(0)=v$. Now, because $\phi (t)$ is a curve in the surface, you have that $f(\phi (t))=0$. So if we let $g(t)=f(\phi (t))$ then $g<\mathrm{quote}/>(t)=0$. Using the chain rule we have that $0=g<\mathrm{quote}/>(0)=(\nabla f)(x).\phi <\mathrm{quote}/>(0)=(\nabla f)(x).v$, which is what we wanted.

There are various ways of looking at this. One is to use Taylor's theorem. Let $(x_0,y_0,z_0)$ be a point on the surface, so $f(x_0,y_0,z_0)=0$.

Then the higher-dimensional form of Taylor's theorem states that

${\displaystyle f(x_0+r,y_0+s,z_0+t)=f(x_0,y_0,z_0)+r\frac{\partial f}{\partial x}(x_0,y_0,z_0)+s\frac{\partial f}{\partial y}(x_0,y_0,z_0)+t\frac{\partial f}{\partial z}(x_0,y_0,z_0)}$

plus some terms of the order of $r$, $s$ and $t$ squared.

You can see that if $(x_0,y_0,z_0)$ is on the surface $f(x,y,z)=0$, then the first term is zero. The second term you can recognise as

${\displaystyle (r,s,t).\nabla f(x_0,y_0,z_0)}$

What does this mean? It means that locally the surface 'looks like' the surface $(x,y,z).\nabla f=$ constant. This is just a plane with normal $\nabla f$. So the normal at that one point $(x_0,y_0,z_0)$ is parallel to $\nabla f$ at that point.

(This stuff is fairly hard, so if there's anything you don't understand, please let me know and I'll explain it more carefully.)