I need to prove that if the distance from a point $P$ contained in the surface $f(x,y,z)=1$ ($f$ is of class $C^1$) to the origin is maximum, then $P$ is perpendicular to the surface, how can I prove this?
Normally I would use Lagrange multipliers with the surface as a constraint, but in this case I don´t even know $f$ so I can´t obtain its partial derivatives, there must be something I'm missing, any help will be appreciated.
You can still use Lagrange multipliers, together with the information that $\nabla f$ is perpendicular to the level surface $\{f = 1\}$. Indeed, it suffices to show that $P$ is parallel to $\nabla f(P)$. To do this we use that $P$ is a solution to the system
\begin{cases} \nabla d^2 = \lambda \nabla f \\ f = 1. \end{cases}
Notice that the first equation evaluated at $P$ yields $2P = \nabla d^2(P) = \lambda\nabla f(P)$, which is exactly what we are after.