Does this statement hold for differentiable functions? I am wondering if this statement hold for differentiable functions.

Assume that $f$ is differntiable. Look at the point $x^*$. For every
  open interval around $x^*$, the tangent at
  $x^*(y=f(x^*)+f'(x^*)(x-x^*))$ has an uncountable number of points in
  common with $f(x)$.
Then $f(x)$ must be straight in an open or half-open interval around
  $x^*$.

Also, does the answer depend on $f$ being differentiable, or just differentiable at $x^*$?
 A: For the second question: Consider the function $f \colon \mathbb{R} \to \mathbb{R}$, 
$$f(x) = \begin{cases} x^2 , \text{ if } x\in \mathbb{Q}, \\
0, \text{ otherwise}, \end{cases} $$
and the point $x^* = 0$. Then $f$ is differentiable at $x^*= 0$ (and only at $x^*)$ and the tangent line at $x^* = 0$ is $y=0$, intersecting the graph in all irrational points on the $x$-axis, in addition to $(0,0)$.
A: Here's a counterexample: Let $b_1>a_1>b_2>a_2 >\cdots \to 0.$ Define
$$A= \bigcup_{n=1}^\infty\, [a_n,b_n].$$
Then define
$$B=\bigcup_{n=1}^\infty\, [-b_n,-a_n].$$
Finally, put $C=A\cup B\cup \{0\}.$ Then $C$ is closed. (It's good to draw a picture. Doing so reveals $C$ is a countable pairwise disjoint union of closed intervals of positive measure sliding in to $0$ from the right, together with intervals symmetric about $0$ of positive measure sliding in to $0$ from the left. We then throw in $\{0\}.$)
Now for any closed subset $E$ of $\mathbb R,$ there exists a function $C^\infty(\mathbb R)$ such that $f=0$ on $E$ and $f>0$ elsewhere. This is a result you will find mentioned many times here on MSE.
So take $E=C$ and consider the function described in the second paragraph. This $f$ is smooth and in every neighborhood of $0,$ $f=0$ on an uncountable set. However if $r>0,$ there are no intervals $[0,r]$ or $[-r,0]$ in which $f\equiv 0.$
