Question: There is a ball drawer.  

Seven color balls will be drawn with the same probability ($1/7$).  

(black, blue, green, yellow, white, pink, orange)

If Anson attempts $9$ times,  

what is the probability that he gets all $7$ different color balls?

My work:

I separate the answer to $3$ ways.

1. $7$ attempts -> done (get $7$ colors)
2. $8$ attempts -> done (get $7$ colors)
3. $9$ attempts -> done (get $7$ colors)

Therefore, my answer is $$\frac{9C7 + 8C7 + 7C7}{7^7 \cdot (7!)}$$
However, I don't know it is correct or not.

Answer: 
The statement is equivalent to
$$\DeclareMathOperator{\arccot}{arccot}
\arccot7+\arccot8=\arccot3-\arccot18
$$
The right-hand side is positive because the arccotangent is decreasing, so it is in the interval $(0,\pi/2)$ and we have to ensure that the same holds for the left-hand side, but this is easy, because $7>1$ and $8>1$, hence
$$
\arccot7+\arccot8<2\arccot1=\frac{\pi}{2}
$$
The two sides are equal if and only if their cotangents are.

Since
$$
\cot(\alpha+\beta)=\frac{\cot\alpha\cot\beta-1}{\cot\alpha+\cot\beta}
\qquad
\cot(\alpha-\beta)=\frac{\cot\alpha\cot\beta+1}{\cot\beta-\cot\alpha}
$$
this amounts to proving that
$$
\frac{7\cdot8-1}{7+8}=\frac{3\cdot18+1}{18-3}
$$
Obviously $\pi+\cot^{-1} 3 \neq \cot^{-1} 3$, since $\pi \neq 0$. Your second method is incorrect because $$\tan^{-1} a+\tan^{-1}b+\tan^{-1}c=\tan^{-1} \left(\frac {a+b+c-abc}{1-ab-bc-ac}\right)$$ is only true for $a,b,c>0$ if $ab+bc+ac<1$.

It is better for you to combine the $\arctan$s two at a time. We have:
$$\tan^{-1}7+\tan^{-1}8=\pi-\tan^{-1} \frac {3}{11} {\tag 1}$$
So $$\tan^{-1}7+\tan^{-1}8+\tan^{-1}18=\left(\pi-\tan^{-1} \frac {3}{11}\right)+\tan^{-1}18=\pi+\left(\tan^{-1}18-\tan^{-1} \frac {3}{11}\right)=\pi+\left(\tan^{-1}\frac {195}{65}\right)=\pi+\tan^{-1} 3$$
Thus we get $$\frac {3\pi}{2}-(\tan^{-1}7+\tan^{-1}8+\tan^{-1} 18)=\frac {3\pi}{2}-(\pi+\tan^{-1}3)=\frac {\pi}{2}-\tan^{-1}3=\cot^{-1}3$$ as expected.

**Note:**

$(1)$: I used the formula $\tan^{-1}a+\tan^{-1}b=\pi+\tan^{-1} \left( \frac {a+b}{1-ab}\right)$ which is true iff $ab>1, a>0, b>0$.

●For calculating $\tan^{-1}18-\tan^{-1}\frac {3}{11}$, I used $\tan^{-1}a-\tan^{-1}b=\tan^{-1}\left(\frac {a-b}{1+ab}\right)$ which is true iff $ab>-1$.