Implicit and parametric curves

Question is: Given a surface in both parametric p(u,v) and implicit function f(p)=0 forms,

Prove that the surface normal of the parametric form is parallel to the normal of the implicit form.

Now I know for implicit it is (df/dx, df/dy, df/dz) and for parametric it is dp/du x dp/dv but I am not sure how to equate them. I think I have to use the chain rule in some way to manipulate the df/dx to get it to equal c*(dy/du dz/dv - dz/du dy/dv) but I’m not sure how.

At a point P, the derivatives along u and v will be zero so
df/du = df/dx dx/du + df/dy dy/du + df/dz dz/du = 0
df/dv = df/dx dx/dv + df/dy dy/dv + df/dz dz/dv = 0
Multiply first by dx/dv and the second by dx/du and subtract to get
df/dy [ dy/du dx/dv - dy/dv dx/du] + df/dz [...] = 0
and do the same again (twice) you can make the argument about [df/dx,df/dy,df/dz] being parallel to the cross product of dx/du and dx/dv

At a point P, the derivatives along u and v will be zero so
df/du = df/dx dx/du + df/dy dy/du + df/dz dz/du = 0
df/dv = df/dx dx/dv + df/dy dy/dv + df/dz dz/dv = 0
Multiply first by dx/dv and the second by dx/du and subtract to get
df/dy [ dy/du dx/dv - dy/dv dx/du] + df/dz [...] = 0
and do the same again (twice) you can make the argument about [df/dx,df/dy,df/dz] being parallel to the cross product of dx/du and dx/dv

Thanks that actually makes perfect sense, the question does further ask: Will the normals remain parallel if the implicit function is also a signed distance function?

My answer would be yes since everything in the solution should still hold, (not entirely sure though)

And further: ‘why is the sign ambiguous?’ And I have no idea why the sign would be ambiguous