This method will evaluate the integral, however not in an basic way using substitution, as I assume you want. I just think some will find this method interesting.
Letting $$f(z)=\frac {z^2}{z^6 + 9}$$
Integrating in the positive sense around the contour formed by a semicircle of radius $R$ (called $C_R$) on the complex plane, we have
$$\int_{C_R}f(z)\, dz=\int_{-R}^R f(z)\, dz+\int_\text{Arc} f(z)\, dz$$
As $R \to \infty$, $\int_\text{Arc} f(z)\, dz=0$, thus
$$\lim_{R \to \infty} \int_{C_R}f(z)\, dz=\int_{-\infty}^\infty f(z)\, dz = 2 \pi i\sum \text{Residues of f(z) in }\lim_{R \to \infty}C_R$$
The poles, $z_1$, $z_2$ and $z_3$ of $f(z)$ are (with de Moivre's theorem)
$$z_1=3^{1/3}\exp(\frac{i \pi}{6})$$ $$z_2=3^{1/3}i$$ $$z_3=3^{1/3}\exp(\frac{5 i \pi}{6})$$
The residues of the poles, $b_1$, $b_2$ and $b_3$ respectively, are (I used L'Hopital's rule and the limit definition of the residue, but an alternative may be possible)
$$b_1=-\frac{i}{18}$$ $$b_2=\frac{i}{18}$$ $$b_3=-\frac{i}{18}$$
So finally,
$$\int_{-\infty}^\infty f(z)\, dz=2\pi i (-\frac{i}{18}+\frac{i}{18}-\frac{i}{18})=-\frac{\pi i^2}{9}=\frac{\pi}{9}$$