Find the coefficient of x3 in the binomial expansion of (3x2/4 - 23x)9.

The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.
In this problem, $a = \frac{3x^2}{4}$, $b = -23x$, and $n = 9$.

We are looking for the coefficient of $x^3$, so we set the exponent $18-r$ equal to 3:

For the binomial expansion $(a+b)^n$, the value of $r$ must be an integer such that $0 \le r \le n$.
In this case, $n=9$, but we found $r=15$. Since $15 > 9$, this value of $r$ is not valid. This means there is no term with $x^3$ in the expansion.

Therefore, the coefficient of $x^3$ is $\boxed{\text{0}}$.