Is there a function $f$ with the following properties:
$x_0$ is a strict local extremum of $f'$.
$(x_0,f(x_0))$ is neither a saddle point of $f$ (i.e. a point with $f'(x_0) =0$ which is not local extremum of $f$) nor an inflection point (i.e. f' changes monotonicy in $x_0$).
On
Let $$f:[-2,1]\rightarrow \mathbb R , \ x\mapsto x^2.$$
Then $f'(x)=2x$ and the point $1$ is a local maximum. We have $f'(x)>0$ for $x\in (0,1]$, thus it is not an infliction point. Also, it isn't a saddle point since $f'(1)\neq 0$.
It is no global extremum since $f(-2)>f(1)$.
I admit, taking the boundary of the domain feels a bit like cheating.
Yes, there is such a function. Define $$f(x) = \begin{cases} x+x^4 + 13x^6\sin (1/x), & x > 0\\x-x^4 + 13x^6\sin (1/x),& x < 0\\ 0,& x =0 \end{cases}$$
Then $f$ is twice continuously differentiable on $\mathbb R.$ Check that there is $a>0$ such that $f'>1$ on $(-a,a)\setminus \{0\},$ and that $f'(0)=1.$ Thus $f'$ has a strict local minimum of $0$ at $0.$ Because $f'(0)\ne 0,$ $(0,0)$ is not a saddle point. Check also that $f''$ takes on positive and negative values in $(0,b)$ for every $b>0.$ The same is true for each $(-b,0).$ Thus $(0,0)$ is not an inflection point.