You have a piecewise defined function here: if $x<9$, then $\min(x,9)=x$ and $f(x)=x^3\cdot x=x^4$. But if if $x>9$, then $\min(x,9)=9$ and $f(x)=x^3\cdot 9=9x^3$.
So, your function is $f(x)=\cases{ x^4, &x<9\cr 9x^3,&x\ge9}$
You'll have to find the derivative separately for each piece:
For $x<9$, \tag{1}f'(x)=(x^4)'=4x^3.
For $x>9$, \tag{2}f'(x)=(9x^3)'=27x^2.
At $x=9$, in order to determine if f'(9) is defined, you need to see if the above formulas "match up" at $x=9$. That is, you need to check that the "derivative at 9" given by (1) and (2) are the same. So, you need to compute the limit of the expression in (1) as $x$ approaches 9 from the left, and the limit of the expression in (2) as $x$ approaches 9 from the right: $ \lim_{x\rightarrow 9^-} 4x^3=4\cdot 9^3. $
$ \lim_{x\rightarrow 9^+}27x^2=27\cdot9^2=3\cdot 9^3. $ Since the two are different, f'(9) is undefined.
See Jonas' astute observation below.
Looking at the graph of $f$ and the expressions (1) and (2), it is not too hard to see that f'(9) exists if and only if the two limits above exist and are equal. The previous statement, however, is not true in general.