Given $\left(4+x^2\right)dy-2x\left(x^2+3y+4\right)dx=0$

$\left(x^2+4\right)\frac{dy}{dx}=2x^3+6xy+8x$

$\left(x^2+4\right)\frac{dy}{dx}-6xy=2x^3+8x$

$\frac{dy}{dx}-\frac{6x}{x^2+4}y=\frac{2x^3+8x}{x^2+4}$

This is of the form of linear differential equation

I.F. $=e^{-\int \frac{6x}{x^2+4}dx}=e^{-3{\log }_e\left(x^2+4\right)}$

$={\mathrm{e}}^{{\log }_e{\left(x^2+4\right)}^{-3}}=\frac{1}{{\left(x^2+4\right)}^3}$

So the solution of the differential equation will be 

$y.\frac{1}{{\left(x^2+4\right)}^3}=\int \frac{2x^3+8x}{{\left(x^2+4\right)}^3\left(x^2+4\right)}dx$

$\frac{y}{{\left(x^2+4\right)}^3}=\int \frac{2x\left(x^2+4\right)}{{\left(x^2+4\right)}^3\left(x^2+4\right)}dx$

Let $x^2+4=t$, $2xdx=dt$

So $\frac{y}{{\left(x^2+4\right)}^3}=\int \frac{dt}{t^3}$

$\frac{y}{{\left(x^2+4\right)}^3}=\frac{-1}{2{\left(x^2+4\right)}^2}+C$

Since the curve passes through origin $\left(0,0\right)$

$0=\frac{-1}{2\times 16}+C\Rightarrow C=\frac{1}{32}$

i.e. $\frac{y}{{\left(x^2+4\right)}^3}=\frac{-1}{2{\left(x^2+4\right)}^2}+\frac{1}{32}$

$y=\frac{-\left(x^2+4\right)}{2}+\frac{{\left(x^2+4\right)}^3}{32}$

Hence $y\left(2\right)=-\frac{8}{2}+\frac{8\times 8\times 8}{32}=12$