Let me try to elaborate so it is more understandable for
@mathstina .
To get the surface of the cone written as a function we need to get all variables on the same side\[z^2=x^2+y^2\implies g(x,y,z)=x^2+y^2-z^2=0\]Now we pick a point (actually a direction vector) on the surface of the cone \(\vec P_0=\langle l,m,n\rangle\)and find the gradient at that point\[\nabla g(x,y,z)=2\langle x,y,-z\rangle\implies \nabla g(l,m,n)=2\langle l,m,-n\rangle\]now let \(\vec P=\langle x,y,z\rangle\) be a point in the tangent plane to the curve at \(\vec P_0\). This implies that the vector \(\vec P-\vec P_0\) is in the tangent plane. Any vector in the tangent plane should be perpendicular to the normal vector (the gradient), so their dot product is zero.\[\nabla g(l,m,n)\cdot(\vec P-\vec P_0)=0\]\[2\langle l,m,-n\rangle\cdot\langle x-l,y-m,z-n\rangle=0\]\[2l(x-l)+2m(y-m)-2n(z-n)=0\]Plugging in the origin to see if it satisfies both the plane and surface equations, at \(x=y=z=0\) we see that the equation of the plane gives\[-2l^2-2m^2+2n^2=0\]\[l^2+m^2-n^2=0\]We can see that this means that these numbers are a solution to \(g(x,y,z)\) and so are on the surface of the cone. Since l, m, and n are arbitrary, this means that every tangent plane to the cone contains the origin.