c[t_]:={Cos[t]-Sin[t],Cos[t],3t}

First we define our own version of the function norm

norm[u_]:=(Sqrt[u.u])

We use the formula K(t)=((||c'(t)×c''(t)||)/(||c'(t)((||)^3)))

((norm[c'[1]\[Cross]c''[1]])/((norm[c'[1]])^3))//N

In the case the ask for the torsion we use τ(t)=((<c'(t),c''(t),c'''(t)>)/(||c'(t)×c''(t)((||)^2)))

((Det[{c'[1],c''[1],c'''[1]}])/((norm[c'[1]\[Cross]c''[1]])^2))//N

X[u_,v_]:={(1+(v^2))Cos[u]-(1+(v^2))Sin[u],(1+(v^2))Cos[u],v}

The unit normal vector is

n=((((∂_u)X[u,v]\[Cross](∂_v)X[u,v])/(norm[(∂_u)X[u,v]\[Cross](∂_v)X[u,v]]))//Simplify)

We write de derivatives of the unit normal vector as a combination of the tangent vectors

coefficientsnu=Solve[(∂_u)n==(Subscript[α, u])(∂_u)X[u,v]+(Subscript[β, u])(∂_v)X[u,v],{(Subscript[α, u]),(Subscript[β, u])}][[1]]//Simplify

coefficientesnv=Solve[(∂_v)n==(Subscript[α, v])(∂_u)X[u,v]+(Subscript[β, v])(∂_v)X[u,v],{(Subscript[α, v]),(Subscript[β, v])}][[1]]//Simplify

We use these coefficientes to obtain the Weingarten transformation

A=((({{(Subscript[α, u]), (Subscript[β, u])}, {(Subscript[α, v]), (Subscript[β, v])}}))/.Join[coefficientsnu,coefficientesnv])

The Gauss curvature is

Det[A]//Simplify

At point (u,v)=(6,6):

Det[A]/.{u->6,v->6}//N

The mean curvature is

(1/2)Tr[A]//Simplify

At point (u,v)=(6,6):

(1/2)Tr[A]/.{u->6,v->6}//N