1. Compute $x$ in terms of $\alpha$

Let $\theta = \tan^{-1}\alpha$. Then $\tan \theta = \alpha$.

Given $x = \sin(2\tan^{-1}\alpha)=\sin 2\theta$ Using $\sin 2\theta = \frac{2\tan\theta}{1+\tan^2\theta}$ we get $x = \frac{2\alpha}{1+\alpha^2}.$

2. Compute $y^2$

Given $y=\sin\left(\frac12\tan^{-1}\frac43\right).$ Let $\phi=\tan^{-1}\frac43.$ Then $\tan\phi=\frac43$.

From a right triangle, $\sin\phi=\frac45,\qquad \cos\phi=\frac35.$ Now use the half-angle formula: $\sin^2\frac\phi2=\frac{1-\cos\phi}{2}.$ So, $y^2=\frac{1-\frac35}{2}=\frac{\frac25}{2}=\frac15.$

3. Use the condition $y^2=1-x$

We have $\frac15=1-x \quad\Rightarrow\quad x=\frac45.$ Since $x=\frac{2\alpha}{1+\alpha^2},$ we solve $\frac{2\alpha}{1+\alpha^2}=\frac45.$ Cross-multiplying: $10\alpha=4(1+\alpha^2)$ $10\alpha=4+4\alpha^2$ $4\alpha^2-10\alpha+4=0$ $2\alpha^2-5\alpha+2=0.$

Factorizing, $2\alpha^2-5\alpha+2=(2\alpha-1)(\alpha-2)=0.$ Thus $\alpha=\frac12 \quad \text{or} \quad \alpha=2.$ So, $S=\left\{\frac12,2\right\}.$

4. Compute the required sum

We need $\sum_{\alpha\in S}16\alpha^3=16\left[\left(\frac12\right)^3+2^3\right].$ Now $\left(\frac12\right)^3=\frac18, \qquad 2^3=8.$ Hence $16\left(\frac18+8\right)=16\left(\frac18+\frac{64}{8}\right)=16\cdot\frac{65}{8}=130.$

5. Final answer

$\boxed{130}$