1. Given recursive definition

We have $f_0(x)=\frac{1}{1-x}$ and $f_{n+1}(x)=f_0(f_n(x))$ for $n=0,1,2,\dots$

We need to find $f_{100}(3)+f_1\left(\frac23\right)+f_2\left(\frac32\right).$

2. Find the pattern of iterates

Let $g(x)=\frac{1}{1-x}.$ Then all $f_n$ are iterates of $g$.

Now compute successive iterates:

Step 1: Compute $g(g(x))$

$g(g(x))=\frac{1}{1-\frac{1}{1-x}}.$ Simplify the denominator: $1-\frac{1}{1-x}=\frac{1-x-1}{1-x}=\frac{-x}{1-x}.$ So, $g(g(x))=\frac{1}{\frac{-x}{1-x}}=\frac{1-x}{-x}=\frac{x-1}{x}.$ Thus, $f_1(x)=g(g(x))=\frac{x-1}{x}.$

Step 2: Compute $g(g(g(x)))$

Now apply $g$ again: $g\left(\frac{x-1}{x}\right)=\frac{1}{1-\frac{x-1}{x}}.$ Simplify: $1-\frac{x-1}{x}=\frac{x-(x-1)}{x}=\frac{1}{x}.$ Hence, $g\left(\frac{x-1}{x}\right)=\frac{1}{1/x}=x.$ So, $f_2(x)=x.$

Therefore the iterates repeat with period $3$: $f_0(x)=\frac{1}{1-x},\qquad f_1(x)=\frac{x-1}{x},\qquad f_2(x)=x,$ and then again $f_3(x)=f_0(x),\ f_4(x)=f_1(x),\ f_5(x)=f_2(x),\dots$

So in general, the sequence is periodic modulo $3$.

3. Evaluate $f_{100}(3)$

Since $100\equiv 1 \pmod 3,$ we get $f_{100}(x)=f_1(x)=\frac{x-1}{x}.$ Hence, $f_{100}(3)=\frac{3-1}{3}=\frac{2}{3}.$

4. Evaluate $f_1\left(\frac23\right)$

Using $f_1(x)=\frac{x-1}{x},$ we get f_1\left(\frac23\right)=\frac{\frac23-1}{\frac23}=rac{-\frac13}{\frac23}=-\frac12.

5. Evaluate $f_2\left(\frac32\right)$

Since $f_2(x)=x,$ we have $f_2\left(\frac32\right)=\frac32.$

6. Add the terms