Perpendicular Vector at Point P on Surface x^3 + xy^3 - z^2 = -4

[ElDavidas]

The question reads:

"Consider the surface given by the equation $$x^3 + xy^3 - z^2 = -4$$ The point p = (1,2,3) lies on this surface. Give a vector that is perpendicular to the surface at p?"

I'm not too confident about this question although there is a theorem in my notes saying:

if p exists within the level surface equal to c and the gradient of the function f is not equal to zero, then the gradient at point p is perpendicular to every path in the level surface equal to c which passes through p.

Does this apply here?

 

[HallsofIvy]

Well, yes, of course, it does! Let f(x,y,z)= x²+ xy³- z². Then this plane is a level curve of f: f(x,y,z)= -4. The gradient of f is perpendicular to that plane.