Given,

$u+iv=\frac{3i}{x+iy+2}$

$\Rightarrow u-iv=\frac{-3i}{x+2-iy}$

$\Rightarrow \left(u+iv\right)\times \left(u-iv\right)=\left\{\frac{3i}{\left(x+2\right)+iy}\right\}\left\{\frac{-3i}{\left(x+2\right)-iy}\right\}$

$\Rightarrow u^2+v^2=\frac{9}{{\left(x+2\right)}^2+y^2}$

$\Rightarrow {\left(x+2\right)}^2+y^2=\frac{9}{u^2+v^2} ...\left(i\right)$

Now, 

$u+iv=\frac{3i}{x+iy+2}$

            $=\left\{\left(\frac{3i}{x+iy+2}\right)\times \left(\frac{x+2-iy}{x+2-iy}\right)\right\}$

             $=\left(\frac{3y+i\left(x+2\right)}{{\left(x+2\right)}^2+y^2}\right)$

              $=\left\{\frac{3y}{{\left(x+2\right)}^2+y^2}\right\}+i\left\{\frac{\left(x+2\right)}{{\left(x+2\right)}^2+y^2}\right\}$

Comparing real and imaginary part, we get

$u=\frac{3y}{{\left(x+2\right)}^2+y^2} ...\left(ii\right)$

By $\left(i\right) \& \left(ii\right)$, we get

$y=\frac{3u}{u^2+v^2}$.