$C:(x,y,z):z=\sqrt{x^2+y^2}$ suppose it is a smooth surface meaning there exists a smooth parametrization
$ φ(u,v):U⊂R^2→C$ such that $φ(u,v)=(φ_1(u,v),φ_2(u,v),φ_3(u,v))$ such that $φ_3=\sqrt{φ_1^2+φ_2^2}$ being continuously differentiable at every point but it fails at the peak since the limit at the peak of the derivative fails to be unique because if i approximate it through different paths(curves) it will be different. (So which is the the derivative( i tried taking the jacobian but not helpfull) and what limit do i take to continue mathematically my proof ).?
Another approach similar. $C:(x,y,z):z=\sqrt{x^2+y^2}$ suppose it is a smooth surface meaning there exists a smooth parametrization
$ φ(u,v):U⊂R^2→C$ now consider the projectio $p:C \rightarrow R^2$. Now i know from a theorem $φ$ is diffeomorpishm also $p\circ φ$ is a diffeomorphism( havent proven why yet) so its inverse is smooth so $φ\circ ((p\circ φ)^{-1}):R^2 \rightarrow C |(x,y,\sqrt{x^2+y^2} )$ is differentiable since both are which is a contradiction