Given,

$\left({\mathrm{x}}^2-4\right)\mathrm{dy}-\left({\mathrm{y}}^2-3\mathrm{y}\right)\mathrm{dx}=0$

$\Rightarrow \int \frac{\mathrm{dy}}{{\mathrm{y}}^2-3\mathrm{y}}=\int \frac{\mathrm{dx}}{{\mathrm{x}}^2-4}$

$\Rightarrow \frac{1}{3}\int \frac{\mathrm{y}-(\mathrm{y}-3)}{\mathrm{y}(\mathrm{y}-3)}\mathrm{dy}=\int \frac{\mathrm{dx}}{{\mathrm{x}}^2-4}$

$\Rightarrow \frac{1}{3}\int \left(\frac{1}{\mathrm{y}-3}-\frac{1}{\mathrm{y}}\right)\mathrm{dy}=\int \frac{\mathrm{dx}}{{\mathrm{x}}^2-2^2}$

$\Rightarrow \frac{1}{3}\left(\log |\mathrm{y}-3|-\log \left|\mathrm{y}\right|\right)=\frac{1}{4}\log \left|\frac{\mathrm{x}-2}{\mathrm{x}+2}\right|+\mathrm{C}$

$\Rightarrow \frac{1}{3}\log \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\frac{1}{4}\log \left|\frac{\mathrm{x}-2}{\mathrm{x}+2}\right|+\mathrm{C}$

At $\mathrm{x}=4, \mathrm{y}=\frac{3}{2}$

$\Rightarrow \frac{1}{3}\log \left|\frac{{\displaystyle \frac{3}{2}}-3}{{\displaystyle \frac{3}{2}}}\right|=\frac{1}{4}\log \left|\frac{4-2}{4+2}\right|+\mathrm{C}$

$\Rightarrow 0=\frac{1}{4}\log \left|\frac{1}{3}\right|+\mathrm{C}$

$\Rightarrow 0=-\log {\left|3\right|}^{\frac{1}{4}}+\mathrm{C}$

$\Rightarrow \mathrm{C}=\frac{1}{4}\log \left(3\right)$

$\Rightarrow \frac{1}{3}\log \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\frac{1}{4}\log \left|\frac{\mathrm{x}-2}{\mathrm{x}+2}\right|+\frac{1}{4}\log \left(3\right)$

Putting, $\mathrm{x}=10$

$\Rightarrow \frac{1}{3}\log \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\frac{1}{4}\log \left|\frac{2}{3}\right|+\frac{1}{4}\log \left(3\right)$

$\Rightarrow \frac{1}{3}\log \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\frac{1}{4}\log \left|2\right|$

$\Rightarrow \log \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\frac{3}{4}\log 2$

$\Rightarrow \log \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\log \left(2^{\frac{3}{4}}\right)$

$\Rightarrow -\mathrm{y}+3=8^{\frac{1}{4}}·\mathrm{y}$

$\Rightarrow 3=\left(8^{\frac{1}{4}}+1\right)y$

$\Rightarrow \frac{3}{1+{\left(8\right)}^{\frac{1}{4}}}=y$