Tangent plane when gradient is zero

Question

I'm attempting to find the tangent plane of the function

$f(x,y,z)=xz+2y^2z^2$ at $(x,y,z)=(-1,1,0).$

The partial derivatives are $z, 4yz^2$, and $4zy^2$ when done in respect to $x, y$ and $z$ respectively. But they're all zero at $(-1,1,0)$ (i.e the gradient is zero).

Does this mean that a tangent plane doesn't exist at this point?

Answer 1

The tangent plane shares the same values of the derivatives of $f$ at that point. As @amd noted, $f_z=-1$. So the tangent plane becomes $t(x,y,z)=f(-1,1,0)-(z-0)=-z$.