Is the tangent plane to the surface $x^4+y^4-z^2=0$ at $(0,0,0)$ undefined? In calculus, the tangent plane to a surface $f(x,y,z)=0$ at $(x_0,y_0,z_0)$ is defined as the plane passing through $(x_0,y_0,z_0)$ with the normal vector $\mathrm{grad} f(x_0,y_0,z_0)$.
According to this description, the tangent plane to the surface $x^4+y^4-z^2=0$ at $(0,0,0)$ should be undefined as the gradient of $x^4+y^4-z^2$ is the zero vector at $(0,0,0)$. However, intuitively speaking, it seems that the plane $z=0$ should be the tangent plane to this surface at $(0,0,0)$:

So in this case, is the plane $z=0$ the tangent plane to the surface $x^4+y^4-z^2=0$ at $(0,0,0)$, or is it undefined since the gradient at $(0,0,0)$ is the zero vector?
 A: The surface is given by $x^{4}+y^{4}-z^{2}=0$ so $z^{2}=x^{4}+y^{4}>0$ so $z=\pm \sqrt{x^{4}+y^{4}}$.
Setting $f(x,y)=\pm \sqrt{x^{4}+y^{4}}$,

*

*$\displaystyle f_{x}(x,y)=\pm \frac{2x^{3}}{\sqrt{x^{4}+y^{4}}}$.

*$\displaystyle f_{y}(x,y)=\pm \frac{2y^{3}}{\sqrt{x^{4}+y^{4}}}$.

Notice that $f\in C^{1}$ because we can extended $f_{x}$ and $f_{y}$ continuously at $(0,0)$ because

*

*$\displaystyle \lim_{(x,y)\to (0,0)}f_{x}(x,y)=0$.

*$\displaystyle \lim_{(x,y)\to (0,0)}f_{y}(x,y)=0$.

as pointed out @Math Lover, so your description says the tangent plane at $P(x_{0},y_{0},z_{0})$ is given by $$z-z_{0}=\nabla f(x_{0},y_{0})\cdot\begin{bmatrix}x-x_{0}\\ y-y_{0}\end{bmatrix}$$
In this case the tangent plane is given by
$$z-0=\begin{bmatrix} 0\\ 0\end{bmatrix}\cdot \begin{bmatrix} x-0\\ y-0\end{bmatrix} \implies z=0 $$
as you intuitively pointed out. I don't know algebraic geometry but by the change of sign "$\pm$" we have the tangent $z^{2}=0$ at $(0,0,0)$ for the surface $x^{4}+y^{4}-z^{2}=0$ as pointed out @Jan-Magnus Økland.
