Foci $(0,\, \pm \sqrt{10}),$ passing through $(2,\,3)$

Here, the foci are on the $y-$ axis.

Therefore, the equation of the hyperbola is of the form $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$

since the foci are $(0,\,\pm \sqrt{10})$,  $c=\sqrt{10}$

We know that $a^{2}+b^{2}=c^{2}$

$\therefore a^{2}+b^{2}=10$

$\Rightarrow b^{2}=10-a^{2}$         ……… $(1)$

since the hyperbola passes through point $(2,\,3)$ 

$\frac{9}{a^{2}}-\frac{4}{b^{2}}=1$         ……… $(2)$

From equations $(1)$ and $(2)$, we obtain

$\frac{9}{a^{2}}-\frac{4}{(10-a)^{2}}=1$

$\Rightarrow 9\left(10-a^{2}\right)-4 a^{2}=a^{2}\left(10-a^{2}\right)$

$\Rightarrow 90-9 a^{2}-4 a^{2}=10 a^{2}-a^{2}$

$\Rightarrow a^{2}-23 a^{2}+90=0$

$\Rightarrow a^{4}-18 a^{2}-5 a^{2}+90=0$

$\Rightarrow a^{2}\left(a^{2}-18\right)-5\left(a^{2}-18\right)=0$

$\Rightarrow\left(a^{2}-18\right)-\left(a^{2}-5\right)=0$

$\Rightarrow a^{2}=18$ or $5$

In hyperbola, $c > a,$ i.e., $c^{2} > a^{2}$

$\therefore a^{2}=5$

$\Rightarrow b^{2}=10-a^{2}=10-5=5$

Thus, the equation of the hyperbola is $\frac{y^{2}}{5}-\frac{x^{2}}{5}=1$