$$f’(x)= \frac 23 (x-2)^{\frac{-1}{3}} (2x+1) + 2(x-2)^{\frac 23}=0$$ $$\implies \frac{2(2x+1)}{3} + 2(x-2)=0$$ $$\implies x=1$$
From the graph, there are actually two points ie. $1,2$. At $x=2$, the function isn’t differentiable, so it can’t be found by this method. Is there any way to obtain $x=2$ algebraicaally?
A critical point doesn't necessarily have to be when the derivative is $0$. It could also be when the function isn't differentiable.
At $x = 2$, the function isn't differentiable, so it is a critical point.