Let the equation of hyperbola be
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\, \ldots\left(i\right)$
Given, $e=\frac{3}{2}$ and foci $=\left(\pm ae, 0\right)=\left(\pm2, 0\right)$
So, $e=\frac{3}{2}$ and $ae=2$
$\Rightarrow a\times\frac{3}{2}=2$
$\Rightarrow a^{2}=\frac{16}{9}$
Also, $b^{2} = a^{2}\left(e^{2}-1\right)$
$\Rightarrow b^{2}=\frac{16}{9}\left(\frac{9}{4}-1\right)=\frac{20}{9}$
On putting the values of $a^{2}$ and $b^{2}$ in $\left(i\right)$, we get
$\frac{x^{2}}{\left(16 / 9\right)}-\frac{y^{2}}{\left(20 / 9\right)}=1$
$\Rightarrow \frac{x^{2}}{4}-\frac{y^{2}}{5}=\frac{4}{9}$