Prove $\nabla f$ is orthogonal to the surface $f$ I'm trying to prove that $\nabla f$ is orthogonal to the surface $f$. I think I have a valid proof but I'm not sure that it is rigorous.
To prove $\nabla f$ is normal  I am proving that $\nabla f\cdot u=0$ for any vector $u$ tangent to the surface $f$. Say $z=f(x,y)$ then $\nabla f$ $= \left(\begin{array}{c}
\delta_x\\
\delta_y\end{array}\right)$ and choosing any vector $v=(v_1,v_2)$ we need to construct the tangent vector to the surface. The tangent vector is $(v_1,v_2,f'_u$(x)).
We achieve $f'_u$(x) from $\nabla f\cdot u$ which gives us $f'_u$(x)$= v_1\delta_x + v_2\delta_y$. Therefore the tangent vector is $u=(v_1,v_2,v_1\delta_x+v_2\delta_y)$. Now we can dot $\nabla f$ with $u$.
However now I came across the problem that we had one vector with 3 rows and one with 2 rows. Therefore I redefined $\nabla f$ and called a new fucntion $g(x,y,z)=f(x,y)-z$ at the level set $g(x,y,z)=0$. Now $\nabla g$ has three components and $$\nabla g=(\delta_x,\delta_y,-1)$$ I then took the dot product of $\nabla g$ and $u$ and this gave $$v_1\delta_x+v_2\delta_y-(v_1\delta_x+v_2\delta_y)=0$$ Therefore $\nabla g$ is orthogonal to the level set $g(x,y,z)=0$ and by equivalence $\nabla f$ is orthogonal to the surface $f(x,y)$. 
Now that is I'm guessing that's a long way round but would you class it as rigorous? Any suggestions to improve it? Is there any way to not have to use $g(x,y,z)$?
Thanks

Consider the level surface $F(x,y,z)=k$. If $\alpha: \mathbb{R} \rightarrow S$ is a curve on the surface then $F(\alpha (t))=k$ for all $t$. But, the chain-rule for multivariate calculus says $\nabla F(\alpha(t)) \cdot \alpha'(t)=0$ for all $t$. Thus, for any time $t_o$ we find the tangent $\alpha'(t_o)$ is perpendicular to $\nabla F(\alpha(t_o))$. But, the curve is arbitrary as is the point and this suffices to show $\nabla F$ is perpendicular to any tangent vector to the level surface $F^{-1}\{k \}$.
