An expansion of order 2 without a second derivative

1. $\left|\frac{f(x)}{x^2}\right|=\left|x\sin\frac1x\right|\le|x|\to0$. Hence $f(x)=0+0\cdot x+0\cdot x^2+o(x^2)$.

2. $\frac{f(x)-f(0)}x=x^2\sin\frac1x\to0$, so $f'(0)=0$. For $x\ne0$,
$$f'(x)=3x^2\sin\frac1x-x\cos\frac1x.$$
Then
$$\frac{f'(x)-f'(0)}x=3x\sin\frac1x-\cos\frac1x,$$
which has no limit as $x\to0$ (the first term tends to $0$ and $\cos\frac1x$ oscillates in $[-1,1]$). So $f''(0)$ does not exist.

Remark. Taylor–Young says: twice differentiable at $a$ $\Rightarrow$ expansion of order $2$. The converse is false for orders $\ge2$ (it is true for order $1$: an expansion of order $1$ is equivalent to differentiability).