Given Foci: $(0,\pm \sqrt {10}) \rightarrow c = \sqrt 10$
Since $a^2+b^2 = c^2 \rightarrow b^2 = \sqrt{10}^2-a^2 = 10-a^2$
Since it passes through $(2,3) \rightarrow \;\large\frac{3^2}{a^2}$$-\large\frac{2^2}{b^2}$$=1$
Substituting for $b^2$, we get $\large\frac{9}{a^2} $$ - \large\frac{4}{10-a^2}$$ = 1$
$\qquad 90 - 9a^2 - 4a^2 - 10a^2 +a^4 = 0 \rightarrow a^4 -23a^2 + 90 = 0$
$\qquad (a^2-18)(a^2-5) = 0$
For a hyperbola, $c \gt a \rightarrow c^2 \gt a^2$.
Since $c^2 = 10, a^2 \neq 18 \rightarrow a^2 = 5$
$\qquad b^2 = c^2 - a^2 = 10 - 5 = 5$
Therefore, the equation of the hyperbola is as follows: $\;\large\frac{y^2}{5}$$-\large\frac{x^2}{5}$$=1$