Answer
$-\dfrac{9x^3}{(x-3)^2}, x\ne-3, 0, 3$
Work Step by Step
The initial restrictions are $x\ne0, 3$.
Use the rule $\frac{a}{b} \div \frac{c}{d}=\frac{a}{b} \cdot \frac{d}{c}$ to obtain:
$\dfrac{3+x}{3-x} \cdot \dfrac{9x^3}{x^2-9}$
$=\dfrac{x+3}{-1(-3+x)} \cdot \dfrac{9x^3}{x^2-3^2}$
$=\dfrac{x+3}{-(x-3)} \cdot \dfrac{9x^3}{x^2-3^2}$
This yields another restriction which is $x\ne -3$
Factor each polynomial completely.
Use special formula $a^2-b^2=(a+b)(a-b)$ to obtain:
$=\dfrac{x+3}{-(x-3)} \cdot \dfrac{9x^3}{(x+3)(x-3)}, x\ne -3, 0, 3$
Cancel the common factor $x+3$:
.
$=-\dfrac{9x^3}{(x-3)^2}, x\ne-3, 0, 3$
Hence, the lowest term is:
$-\dfrac{9x^3}{(x-3)^2}, x\ne -3, 0, 3$.
