1. Differentiate both functions

Given $f(x)=2x+\tan^{-1}x,$ so $f'(x)=2+\frac{1}{1+x^2}.$

Also, $g(x)=\ln\big(\sqrt{1+x^2}+x\big).$ This is a standard form: $\ln\big(x+\sqrt{1+x^2}\big)=\sinh^{-1}x,$ therefore $g'(x)=\frac{1}{\sqrt{1+x^2}}.$

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2. Check option A

We need to see whether there exists $\hat x\in[0,3]$ such that $f'(\hat x)<g'(\hat x).$

Now for $x\in[0,3]$, $f'(x)=2+\frac{1}{1+x^2}>2,$ while $g'(x)=\frac{1}{\sqrt{1+x^2}}\le 1.$ Hence for all $x\in[0,3]$, $f'(x)>g'(x).$ So such $\hat x$ does not exist.

Therefore, A is false.

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3. Check option B

We test whether there exist $0<x_1<x_2<3$ such that $f(x)<g(x)\quad \forall x\in(x_1,x_2).$

Consider $h(x)=f(x)-g(x).$ Then $h'(x)=f'(x)-g'(x)=2+\frac{1}{1+x^2}-\frac{1}{\sqrt{1+x^2}}.$

Let $t=\sqrt{1+x^2}\ge 1$. Then $\frac{1}{1+x^2}=\frac{1}{t^2},$ so $h'(x)=2+\frac{1}{t^2}-\frac{1}{t}.$ Since $t\ge 1$, $2+\frac{1}{t^2}-\frac{1}{t}>0.$ Hence $h'(x)>0$ for all $x\in[0,3]$, so $h$ is strictly increasing.

Now $h(0)=f(0)-g(0)=0-\ln 1=0.$ Since $h$ is increasing and $h(0)=0$, $h(x)>0\quad \forall x>0.$ Thus $f(x)>g(x)\quad \forall x\in(0,3].$ So there cannot be any interval where $f(x)<g(x)$.

Therefore, B is false.

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4. Check option C

We compute $\min f'(x)$ and $\max g'(x)$ on $[0,3]$.

For $f'(x)$:

$f'(x)=2+\frac{1}{1+x^2}.$ Since $\frac{1}{1+x^2}$ decreases on $[0,3]$, $f'(x)$ decreases. Thus $\min f'(x)=f'(3)=2+\frac{1}{10}=\frac{21}{10}.$

For $g'(x)$:

$g'(x)=\frac{1}{\sqrt{1+x^2}},$ which also decreases on $[0,3]$. Thus $\max g'(x)=g'(0)=1.$

Then $1+\max g'(x)=1+1=2.$ But $\min f'(x)=\frac{21}{10}\ne 2.$ So the statement is false.

Therefore, C is false.

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5. Check option D

Since $f'(x)>0 \quad \text{and} \quad g'(x)>0$ for all $x\in[0,3]$, both functions are increasing on $[0,3]$. Therefore, $\max f(x)=f(3), \qquad \max g(x)=g(3).$

Now $f(3)=2\cdot 3+\tan^{-1}3=6+\tan^{-1}3.$ Also $g(3)=\ln(\sqrt{10}+3).$

Numerically, $\tan^{-1}3\approx 1.249,$ so $f(3)\approx 7.249.$ And $\sqrt{10}+3\approx 6.162,$ so $g(3)=\ln(6.162)\approx 1.818.$ Hence $\max f(x) > \max g(x).$

Therefore, D is true.

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6. Final conclusion

Only option D is correct.