I want to find a function $f:\Bbb{R}^n\rightarrow\Bbb{R}$ such that:
$1)$ $f(0)=f'(0)=0$;
$2)$ exists $C>0$ and a neighborhood $U$ of $0\in\Bbb{R}^n$ such that $|f(x)|\leq C|x|^2$
$3)$ there is no $K>0$ such that $$\left\vert\dfrac{\partial^2 f}{\partial x_i x_j(x)}\right\vert\leq K$$ for all $x$ with $|x|<1$ and $i,j\in\{1,\dots n\}.$
I started with the function $f(x_1,\dots, x_n)=\dfrac{2}{3}x_1^{3/2}$. I have
$$f'(x)=\left(\sqrt{x},0,\dots 0\right) \implies f'(0)=0 $$
Furthermore, $$\dfrac{\partial^2f}{\partial x_i^2}(x)=\dfrac{1}{2\sqrt{x}},$$
which is unbounded in $B_{1}(0)\subset \Bbb{R}^{n}$, so $f$ satisfies the conditions $1$ and $3$.
However, $|f(x)|=\dfrac{2}{3}|x_1|^{3/2}\leq |x|...$ and I don't know if $f$ will satisfy condition $2)$.
Am I in the right way to find this function?
Hint: Your function $\ f(x)=\frac{2}{3}x_1^\frac{2}{3}\ $ doesn't satisfy condition $(2)$, because $\ \left|\frac{f(x)}{|x|^2}\right|=\left|\frac{2}{3\sqrt{x}_1}\right|\rightarrow0\ $ as $\ x\rightarrow0\ $.
I suggest you look at something like $$ f(x)=\cases{0&if $\ x=0\ $\\ \int_\limits{0}^{\hspace{1em}x_1}y^2\sin\left(\frac{1}{\left|y\right|^n}\right)dy&otherwise,} $$ for some suitably chosen $\ n\ $.