The partial fraction of $\frac{1}{\left(x^2+9\right)\left(x^2+16\right)}$  are

$1=\left(Ax+B\right)\left(x^2+16\right)+\left(Cx+D\right)\left(x^2+9\right)$

$A+C=0\to \left(1\right);B+D=0\to \left(2\right);16A+9C=0\to \left(3\right)$ $;16B+9D=1\to \left(4\right)$

From (1) and (3)

$A=0,C=0$

From (2) and (4)

$B=\frac{1}{7},D=\frac{-1}{7}$

$\frac{1}{7\left(x^2+9\right)}-\frac{1}{7\left(x^2+16\right)}$

$=\frac{1}{7}\left(\frac{1}{x^2+9}-\frac{1}{x^2+16}\right)$