1. Analyze each piece of $f(x)$

We are given

$$f(x)= \begin{cases} x^5+5x^4+10x^3+10x^2+3x+1, & x<0 \\ x^2-x+1, & 0\le x<1 \\ \dfrac23 x^3-4x^2+7x-\dfrac83, & 1\le x<3 \\ (x-2)\ln(x-2)-x+\dfrac{10}{3}, & x\ge 3 \end{cases}$$

We check the options one by one.

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2. Option A: $f$ is increasing on $(-\infty,0)$

For $x<0$,

$$f(x)=x^5+5x^4+10x^3+10x^2+3x+1.$$

Differentiate:

$$f'(x)=5x^4+20x^3+30x^2+20x+3.$$

Notice

$$5x^4+20x^3+30x^2+20x+3=5(x+1)^4-2.$$

So for $x<0$,

$$f'(x)=5(x+1)^4-2.$$

Now $f$ will be increasing on $(-\infty,0)$ only if $f'(x)>0$ for all $x<0$. But at $x=-1$,

$$f'(-1)=5(0)^4-2=-2<0.$$

Hence $f$ is not increasing on all of $(-\infty,0)$.

So, A is false.

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3. Option B: $f'$ is not differentiable at $x=1$

First check continuity and differentiability of $f$ at $x=1$.

Left side ($0\le x<1$):

$$f(x)=x^2-x+1$$ $$f'(x)=2x-1$$

So,

$$f'_-(1)=2(1)-1=1.$$

Also,

$$f(1^-)=1^2-1+1=1.$$

Right side ($1\le x<3$):

$$f(x)=\frac23 x^3-4x^2+7x-\frac83$$ $$f'(x)=2x^2-8x+7$$

So,

$$f'_+(1)=2-8+7=1.$$

Also,

f(1)=\frac23-4+7-\frac83=rac{2-8+21-8}{3}=1.

Thus $f$ is differentiable at $x=1$ and

$$f'(1)=1.$$

Now check differentiability of $f'$ at $x=1$, i.e. compare second derivatives from both sides.

• For $0\le x<1$,

$$f'(x)=2x-1 \implies f''(x)=2.$$

So,

$$(f')'_-(1)=2.$$

• For $1\le x<3$,

$$f'(x)=2x^2-8x+7 \implies f''(x)=4x-8.$$

So,

$$(f')'_+(1)=4(1)-8=-4.$$

Since left and right derivatives of $f'$ at $x=1$ are different, $f'$ is not differentiable at $x=1$.

So, B is true.

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4. Option C: $f$ is onto

We need to check whether range of $f$ is all of $\mathbb R$.

Consider the first piece for $x<0$:

$$f(x)=x^5+5x^4+10x^3+10x^2+3x+1.$$

As $x\to -\infty$, dominant term is $x^5$, so

$$f(x)\to -\infty.$$

Also,

$$\lim_{x\to 0^-} f(x)=1.$$

Since this branch is a polynomial, it is continuous on $(-\infty,0)$, so its range contains all values from very large negative numbers up to some values near $1$. In particular, it gives all sufficiently negative real numbers.

Now examine the remaining pieces:

On $[0,1)$:

$$f(x)=x^2-x+1=\left(x-\frac12\right)^2+\frac34.$$

Hence range here is

$$\left[\frac34,1\right].$$

On $[1,3)$:

$$f(x)=\frac23 x^3-4x^2+7x-\frac83.$$

At $x=1$,

$$f(1)=1.$$

At $x=2$,

$$f(2)=\frac{16}{3}-16+14-\frac83=\frac{8}{3}-2=\frac23.$$

At $x\to 3^-$,

$$f(3^-)=18-36+21-\frac83=3-\frac83=\frac13.$$

So this piece gives values down to $\frac13$.

On $[3,\infty)$:

$$f(x)=(x-2)\ln(x-2)-x+\frac{10}{3}.$$

At $x=3$,

$$f(3)=1\cdot \ln 1-3+\frac{10}{3}=\frac13.$$

Differentiate:

$$f'(x)=\ln(x-2).$$

Thus for $x>3$, $f'(x)>0$, so this branch is increasing. Also,

$$\lim_{x\to \infty} \big((x-2)\ln(x-2)-x+\tfrac{10}{3}\big)=\infty.$$

Therefore this branch gives range

$$\left[\frac13,\infty\right).$$

Combining, the last branch alone already gives all values from $\frac13$ to $\infty$. The first branch gives all values tending to $-\infty$ and approaching $1$ near $0^-$. Since it is continuous and $f(-1)=2$, $f(-2)=-1$, it certainly covers all values below $1$, including all negative values and values in $( -\infty,1 ]$ through continuity on intervals where it crosses them. In particular, together with $[\frac13,\infty)$ from later pieces, every real value is attained.

Hence, $f$ is onto.

So, C is true.

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5. Option D: $f'$ has a local maximum at $x=1$

Compute $f'$ on both sides of $1$:

• For $0\le x<1$,

$$f'(x)=2x-1.$$

This is increasing, and as $x\to 1^-$,

$$f'(x)\to 1.$$

Also for $x<1$, we have $f'(x)<1$.

• For $1\le x<3$,

$$f'(x)=2x^2-8x+7.$$

At $x=1$,

$$f'(1)=1.$$

For $x>1$ close to $1$,

$$f''(x)=4x-8<0$$

(since near $1$, this is negative), so $f'(x)$ decreases to the right of $1$. Hence for $x>1$ sufficiently close to $1$, $f'(x)<1$.

Therefore, in a neighborhood of $x=1$,

$$f'(x)\le f'(1),$$

with strict inequality for $x\ne 1$ close enough to $1$. So $f'$ has a local maximum at $x=1$.

So, D is true.

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6. Final conclusion

• A: False
• B: True
• C: True
• D: True

Thus the correct options are

$$\boxed{B,\ C,\ D}$$

This matches the stored correct answer.