1. We need to study differentiability of $(f\circ g)(x)=f(g(x)).$

Given $f(t)=\begin{cases} t+2, & t<0 \\ t^2, & t\ge 0 \end{cases}$ and $g(x)=\begin{cases} x^3, & x<1 \\ 3x-2, & x\ge 1. \end{cases}$

The composition can fail to be differentiable at points where:

1. $g$ is not differentiable,
2. $g(x)$ hits the switching point of $f$, i.e. $g(x)=0$.

So we check both.

2. First, check differentiability of $g$.

For $x<1$, $g(x)=x^3$. For $x\ge 1$, $g(x)=3x-2$.

At $x=1$: $g(1^-)=1^3=1, \qquad g(1)=3(1)-2=1.$ So $g$ is continuous at $x=1$.

Now derivatives: $g'_-(1)=3x^2\big|_{x=1}=3, \qquad g'_+(1)=3.$ Hence $g$ is differentiable at $x=1$.

Therefore, $g$ is differentiable for all $x\in\mathbb R$.

3. Now check where $f$ is differentiable.

For $t<0$, $f(t)=t+2$. For $t\ge 0$, $f(t)=t^2$.

Possible issue is at $t=0$.

Continuity at $t=0$: $\lim_{t\to 0^-} f(t)=\lim_{t\to 0^-}(t+2)=2,$ $f(0)=0^2=0.$ Since $2\ne 0$, $f$ is not continuous at $t=0$, hence not differentiable at $t=0$.

For all $t\ne 0$, $f$ is differentiable.

So $(f\circ g)(x)$ can only fail to be differentiable when $g(x)=0.$

4. Solve $g(x)=0$.

Case 1: $x<1$. Then $g(x)=x^3$, so $x^3=0 \implies x=0.$ Since $0<1$, this is valid.

Case 2: $x\ge 1$. Then $g(x)=3x-2$, so $3x-2=0 \implies x=\frac23.$ But $\frac23<1$, so this is invalid for this branch.

Thus the only point where $g(x)=0$ is $x=0.$

5. Verify non-differentiability at $x=0$ by writing the composition explicitly.

First determine sign of $g(x)$ near $x=0$. For $x$ near $0$ (and $x<1$), $g(x)=x^3.$ So:

• if $x<0$, then $g(x)<0$, hence $f(g(x))=g(x)+2=x^3+2$,
• if $x\ge 0$ and near $0$, then $g(x)\ge 0$, hence $f(g(x))=(g(x))^2=x^6$.

Therefore near $x=0$, $(f\circ g)(x)=\begin{cases} x^3+2, & x<0, \\ x^6, & x\ge 0. \end{cases}$

At $x=0$, $\lim_{x\to 0^-}(f\circ g)(x)=2, \qquad (f\circ g)(0)=0.$ So it is discontinuous at $x=0$, hence not differentiable there.

6. Check if there are any other problematic points.

• At $x=1$, $g$ is differentiable and $g(1)=1\ne 0$, and $f$ is differentiable at $1$. So $(f\circ g)$ is differentiable at $x=1$.
• For all other $x$, both $g$ and $f$ at $g(x)$ are differentiable.

Hence there is exactly one point where $(f\circ g)(x)$ is not differentiable.

7. Final answer: $\boxed{1}$ So the correct option is C.