How can we show that it is perpendicular to the tangent plane? We consider the surface $S$ defined by $f(x, y, z) = 0$, where $f$ is a smooth function such that $\nabla f$ does not vanish at any point of $S$. 
How can we show that $\nabla f$ is perpendicular to the tangent plane at every point of $S$, and deduce that $S$ is orientable? 

I assume $f:\mathbb{R}^3\to\mathbb{R}$. Let $x\in S$, then the tangent space of $S$ at $x$ can be seen as the space of velocity vectors at $x$ of smooth curves (passing through $x$, of course). So let $\gamma:I\to S$ be a smooth curve. Can you show that $\langle\dot{\gamma},\nabla f\rangle = 0$?
