If $\tan \left(\frac{\pi }{4}+\frac{y}{2}\right)={\tan }^3\left(\frac{\pi }{4}+\frac{x}{2}\right),\text{ then }\frac{3+{\sin }^{2}x}{1+3{\sin }^{2}x}=$

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

Squaring on both sides $\frac{1+\sin y}{1-\sin y}={\left(\frac{1+\sin x}{1-\sin x}\right)}^3$

Apply componendo and dividendo rule

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

$\Rightarrow \frac{3+{\sin }^{2}x}{1+3{\sin }^{2}x}=\frac{\sin y}{\sin x}$