Answer
$$\frac{\left(v-1\right)\left(v^2+9\right)}{v-3}$$
Work Step by Step
Recall, one never divides by a fraction. Rather, if we want to divide by a fraction, we multiply by the reciprocal of the fraction. Recall, the reciprocal of a fraction is that fraction with the numerator and the denominator switched. Thus, we find:
$$ \frac{\left(v^2-1\right)\left(v^2+9\right)}{\left(v+1\right)\left(v-3\right)}\\ \frac{\left(v+1\right)\left(v-1\right)\left(v^2+9\right)}{\left(v+1\right)\left(v-3\right)}\\ \frac{\left(v-1\right)\left(v^2+9\right)}{v-3}$$