Prove, by geometric or algebraic arguments, that $df^{'}$, the gradient of dual space of R3, is normal to surfaces of constant f.
Being the first time i am seeing things like this, is a little confused to me yet to know if i am doing right or not. See:
The metric in R3 Euclidean space is just the identity $\delta_{a}^{a}$, in such way that the componentes of the "dual gradient (i am saying dual gradient because, as far as i know, there is the vector gradient too)" are equal to the gradient in the "original" space. So, since the vector gradient is normal to surfaces of constant f, so is its dual.
A good explanation is the following. If you have a curve $M(t)$ on the surface $f= C$, then $f(M(t)=C$. Using the chain rule $< \nabla f, M'(t)>=0$. The gradient and the tangent vectors are perpendicular.