$y =\begin{cases} 2 - 3x,\quad \quad \quad 2 - 3x &\geq 0 \\
3x - 2, \quad \quad \quad 2 - 3x &< 0 \end{cases}$
$\implies y = \begin{cases}2 - 3x, \quad \quad \quad x &\leq \frac{2}{3}\\
3x - 2, \quad \quad \quad x &> \frac{2}{3}\end{cases}$
As $y$ is polynomial it is continuous and differentiable at all points but do not know at $x = \frac{2}{3}$.
To check continuity at $x = \frac{2}{3}$
Left limit = $2-3 \times \frac{2}{3} = 0$
Right limit = $ 3 \times \frac{2}{3} - 2 = 0$
$f(a) = f(2/3) = 2-3 \times \frac{2}{3} = 0$
$\because \text{LL = RL} = f(a), y\text{ is continuous } \forall_{x \epsilon R}.$
To check differentiability at $x = \frac{2}{3}$
Left derivative $= 0 - 3 = -3$
Right derivative $= 3 - 0 = 3$
$\because \text{LD} \neq \text{RD}, y\text{ is not differentiable at }x = \frac{2}{3}.$
So, Answer is option C.