Given $f\left(0\right)=0, f\left(1\right)=1$ and $f\left(2\right)=2$

Let, $h\left(x\right)=f\left(x\right)-x$

Clearly $h\left(x\right)$, will be continuous and twice differentiable on $\left(0, 2\right)$.

Now, $h\left(0\right)=h\left(1\right)=h\left(2\right)=0$

By Rolle's mean value theorem in $\left[0, 1\right]$, we get

$h^{\text{'}}\left(c_1\right)=0$

$\Rightarrow f^{\text{'}}\left(c_1\right)-1=0$

$\Rightarrow f^{\text{'}}\left(c_1\right)=1$, where $c_1\in \left(0, 1\right)$

Also, in the interval $\left[1, 2\right]$

$h^{\text{'}}\left(c_2\right)=0$

$\Rightarrow f^{׀}\left(c_2\right)-1=0$

$\Rightarrow f^{\text{'}}\left(c_2\right)=1$, where $c_2\in \left(1, 2\right)$

Now, use Rolle's theorem on $\left[c_1, c_2\right]$ for $h^{\text{'}}\left(x\right)$, 

 We have, $h^{\text{'}\text{'}}\left(c\right)=0$

$\Rightarrow f^{\text{'}\text{'}}\left(c\right)=0$, where $c\in \left(c_1, c_2\right)$

Hence,  $f^{\text{'}\text{'}}\left(c\right)=0$ for some $x\in \left(0, 2\right)$.