option D) is correct.
$f(x)=x^2 , x>=0$
$=-x^2 , x<0$
at $x=0$ function f(x) is continuous since $LHL=RHL=f(0) = 0$
now, first derivative of f(x)
$f'(x) = 2x , x>=0$
$=-2x , x<0$
since, $f'(x=0^+)=0$ and $f'(x=0^-)=0$ i.e both are equal,
function f(x) is diffrentiable at $x=0$
derivative of first derivative,
$f''(x) = 2 , x>=0$
$ =-2 , x<0$
since $f''(x=0^+) = 2$ and $f''(x=0^-) = -2$ i.e both are unequal
derivative of f(x) is not diffrentiable at $x=0$