Let $S\subset R^3$ a surface described by the function $f(x,y,z)=1$ where $f$ is a $C^1$ function.Suppose that $P$ is farthest point from the surface to the origin.
Prove that the vector $\vec {OP}$ is normal to the surface $S$
The first think that came to my mind was lagrange multipliers: so i need to maximize $g(x,y,z)=\sqrt{x^2+y^2+z^2}$ subject to $f(x,y,z)-1=0$
Then $\nabla g(x,y,z)=\lambda \nabla f$ then $$({x\over \sqrt{x^2+y^2+z^2}}, {y\over \sqrt{x^2+y^2+z^2}}, {z\over \sqrt{x^2+y^2+z^2}})=\lambda({\partial f\over \partial x},{\partial f\over \partial y},{\partial f\over \partial z})$$
How can I proceed from here do I need to solve the differential equations? Any ideas would be appreciated
First, you don't have to solve a differential equation. The equation you found is valid only at some points.
However, with what you did, you're almost done. If $\vec{n}=(u,v,w)$ is a unit vector normal to the surface S at a point $P=(x,y,z)$ of the surface, you have $$\begin{cases} f(x,y,z)=1 & \text{ as the point is supposed to belong to S}\\ \vec{n}.\nabla f = 0 & \text{ as the vector is normal to the surface} \end {cases}$$ As you noticed, $\vec{OP}$ is colinear to $\nabla f$ (at point $(x,y,z)$). Hence you also have $$\vec{OP}.\vec{n}=0$$ which is the desired result.