I've a function $f=f(x,y,z)$ and a given constant vector $\vec A$.
Then, there is an equation $\nabla f \cdot \vec A=\rm something$ that for me has a physical meaning.
I'd like to write the same equation in the case in which there is the constraint $x=yz$ and $f=x^2+y^2+z^2$ .
My problem is that I don't know how express the term $\nabla f \cdot \vec A$ in this case
Constraining $f$ doesn't affect its gradient, if you still allow $f$ to exist in three dimensional space. You could write a new function $g(y,z)=(yz)^2+y^2+z^2$ but this is now a two dimensional function, with two dimensional gradient, but its gradient would not agree with $f$'s gradient because it is in a different domain. Constraining just asks the question: what does $f$ look like in the domain where $x=yz$? But the gradient, its direction of steepest ascent, never changes - the constraint just means that a step in the direction of steepest ascent may be impossible, outside of the constraint. I suggest you research Lagrange multipliers for a demonstration of gradients and equality constraints; you'll see that we do not adjust the gradient of $f$ at all. The quantity $\nabla f\cdot\vec A$ is still the same! The function $f$ itself is unchanged, and the constraint merely yields the set of all points $f(yz,y,z)$ and does not yield the new function $g$!
You could perhaps visualise this as $f$ walking in the boundary surface, where $x=yz$, and the gradient stays unchanging, at times perhaps "tugging" $f$ outside of the constrained domain, at other times following it,... $\nabla f$ still acts in the same ways.