$f'$ is bounded and isn't continuous on $(a,b)$, so there's a point $y\in(a,b)$ such that $\lim_{x\to y}f'$ does not exist 
Prove/disprove: $f$ has a bounded derivative and $f'$ isn't continuous on $(a,b)$, so there's a point $y\in(a,b)$ such that $\displaystyle\lim_{x\to y}f'$ does not exist.

I think that if $f'$ isn't continuous on the interval, then maybe we could have two disjoint sub-intervals, like for example $(a,c), (d,b)$ such that $d-c=\dfrac {a+b} 3$ so there's a substantial gap in the interval $(a,b)$ where $f'$ isn't defined so it follows that it won't have a limit there, for example: on $\dfrac {c+d}2$. 

The very fundamental thing one needs to observe here is that a derivative can't have jump discontinuity. If $f'(x) \to L$ as $x \to c$ then $f'(c) = L$ and thus $f'$ is continuous at $c$. Hence it is not possible for a derivative to have a limit at point and not to be continuous at that point. It follows that there will be points where limit of derivative $f'(x)$ does not exist.
Now to the proof of the statement mentioned in bold above. One proof is already given by leshik in his answer. Here is another proof based on mean value theorem. Suppose that $f(x)$ is differentiable in neighborhood of $c$. Suppose $x \to c$ and $f'(x) \to L$ as $x \to c$. Then we have $f(x) - f(c) = (x - c)f'(d)$ where $d$ lies somewhere between $c$ and $x$. Thus $$\frac{f(x) - f(c)}{x - c}=f'(d)$$ When $x\to c$ then LHS tends to $f'(c)$ and RHS tends to $L$ because $d \to c$. It follows that $f'(c) = L = \lim_{x \to c}f'(x)$. Hence $f'(x)$ is continuous at $c$.
