Let $S$ be the surface that is the graph of a continuous function $f: U \rightarrow \mathbb{R}$ on an open $U \subset \mathbb{R}^2$. Let $p = (x, y, f(x, y)) \in S$.
One usually defines the tangent plane $P$ to $S$ at $p$ in the situation that the function $f$ be differentiable at $(x,y)$ — in the usual technical sense. [And, in particular, in the case that $f$ has continuous first-order partial derivatives in some neighborhood of $(x, y)$.]
A consequence of differentiability at $(x,y)$ is the following property:
For every smooth curve $\gamma: I \rightarrow S$ (on some open interval) through $p$, the tangent line to $\gamma(I)$ at $p$ lies in the plane $P$.
Question: Is the existence of a plane $P$ through $p$ having that property sufficient for $f$ to be differentiable at $(x, y)$?
[Of course it is not sufficient to consider just curves cut out by planes through $p$ and perpendicular to the $(x,y)$-plane: the directional derivative of $f$ at $(x, y)$ in every direction may exist yet $f$ need not be differentiable there.]
Here is a specific counterexample, not quite as strong as what I said in my comment but much easier to formulate.
Let $S$ be the cone $z=\sqrt{x^2+y^2}$. This is just the graph of $f(x,y) = \sqrt{x^2+y^2}$. This function $f$ has no directional derivative in any direction (although it does have "one-sided directional derivatives"). The graph of $f$ is the cone $S$ described above, and $S$ has no tangent plane at the cone point $p=(0,0,0)$. And yet, any plane through $(0,0,0)$ vacuously satisfies the definition in the box, because there do not exist any smooth curves on $S$ through $p$ possessing a tangent line at $p$.
If you want a stronger example where there do not even exist any "one-sided directional derivatives" at $(0,0)$, take $f(x,y) = \alpha\bigl(\sqrt{x^2+y^2}\bigr)$ where $$\alpha(r)= \begin{cases} 0 & r=0 \\ r \sin(1/r) & r > 0 \end{cases} $$