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Let $\mathbf{c}\in E\subseteq\Bbb R^2$, $f:E\to\Bbb R$ and $f$ is $C^1$ at $\mathbf{c}$. I want to prove that all tangent lines through $\mathbf{c}$ indeed forms a plane. In a lengthier but clear sentence, "all the tangent vectors of arbitrary curve on the graph of $f$ through $\mathbf{c}$ form a plane"$(\star)$.

I know that in the elementary calculus course, the way the author to prove that $f$ has a tangent plane at $\mathbf{c}$, is usually begin by first finding the tangent plane of the level surface form $F(x,y,z)=C$(the process involved using the chain rule) and then rewrite $f(x,y)$ as this form($z-f(x,y)=0$), applying the former result to get the equation of the tangent plane, and then one have shown that such $f$ has a tangent plane at $\mathbf{c}$, which means $(\star)$.

But now, since intuitively we all know that, let $\mathbf u=(u_1,u_2)\in\Bbb R^2$ be any unit vector, then the sets of the vectors $(u_1,u_2,D_\mathbf{u}f(\mathbf{c}))$ made by the directional derivative on all direction $\mathbf u$ is the all possibility that the tangent vectors of arbitrary curve on the graph of $f$ through $\mathbf{c}$ could be. So if we turn to prove that for all $\mathbf{u}$, $(u_1,u_2,D_\mathbf{u}f(\mathbf{c}))$ is contained in the same plane, then it is suffice to show that $f$ has a tangent plane at $\mathbf{c}$ right? I think this way is more direct, how can I finish this?

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    The vector product of two vectors you mentioned will have to be perpendicular all your vectors. – 2017-02-06
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    Thanks! Can I avoid the cross product? Because I also want it to be able to extend to higher dimensions. – 2017-02-06
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    We can combine the two perspectives: We know that the graph of $f$ is a level surface of $F(x, y, z) := z - f(x, y)$ and that vectors in the tangent plane at ${\bf c} = (x_0, y_0, z_0)$ will be precisely those orthogonal to $(\nabla F)({\bf c}) = (-f_x(x_0, y_0), -f_y(x_0, y_0), 1)$. Now, compute $(\nabla F)({\bf c}) \cdot (u_1, u_2, (D_{\bf u} f)(x_0, y_0))$ (like you say, by linearity we may as well just verify this for unit $\bf u$). This extends to graphs of functions of more variables formally. – 2017-02-06

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Because $f$ is $C^1$, $f$ is differentiable at $\mathbf c$ with derivative a linear map $Df_{\mathbf c}\colon\Bbb R^2\to\Bbb R$. Each tangent line is of the form $T(\mathbf v) = \big(\mathbf v,Df_{\mathbf c}(\mathbf v)\big)$ for some $\mathbf v\in\Bbb R^2$. The collection of all such is the image of the linear map $T\colon\Bbb R^2\to\Bbb R^3$, and this is, of course, the tangent plane of the surface at $\mathbf c$.

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    Thanks. I can't figure out why each tangent line is of the form $T(\mathbf v) = \big(\mathbf v,Df_{\mathbf c}(\mathbf v)\big)$. – 2017-02-07
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    Because the directional derivative in direction $\mathbf v$ is precisely $Df_{\mathbf c}(\mathbf v)$. When $\mathbf v$ is a unit vector, this is the slope of the tangent line headed in direction $\mathbf v$. – 2017-02-07
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    I see! Thanks! :) – 2017-02-07