Showing whether two numbers are equal or not $\dfrac{\sin (2x+y)}{\sin (2x)} =\dfrac{\sin (x+2y)}{\sin (2y)}$,where $0<x,y\le\dfrac{\pi}{4}$ .
Can I show that $x=y $ or find two numbers $x,y$ such that $x\not=y$?

Start with
$$
\frac{\sin (2x+y)}{\sin (2x)} =\frac{\sin (x+2y)}{\sin (2y)}\tag{1}
$$
Regrouping $(1)$, we get
$$
\frac{\sin\left(\frac{3}{2}(x+y)+\frac{1}{2}(x-y)\right)}{\sin\left((x+y)+(x-y)\right)}=\frac{\sin\left(\frac{3}{2}(x+y)-\frac{1}{2}(x-y)\right)}{\sin\left((x+y)-(x-y)\right)}\tag{2}
$$
Expanding $(2)$, yields
$$
\begin{align}
&\frac{\sin\frac{3}{2}\!\!(x+y)\;\cos\frac{1}{2}\!\!(x-y)+\cos\frac{3}{2}\!\!(x+y)\;\sin\frac{1}{2}\!\!(x-y)}{\sin(x+y)\;\cos(x-y)+\cos(x+y)\;\sin(x-y)}\\
&=\frac{\sin\frac{3}{2}\!\!(x+y)\;\cos\frac{1}{2}\!\!(x-y)-\cos\frac{3}{2}\!\!(x+y)\;\sin\frac{1}{2}\!\!(x-y)}{\sin(x+y)\;\cos(x-y)-\cos(x+y)\;\sin(x-y)}\tag{3}
\end{align}
$$
Since $\frac{a+b}{c+d}=\frac{a-b}{c-d}\Rightarrow\frac{b}{a}=\frac{d}{c}$, $(3)$ implies
$$
\frac{\tan\frac{1}{2}\!\!(x-y)}{\tan\frac{3}{2}\!\!(x+y)}=\frac{\tan(x-y)}{\tan(x+y)}\tag{4}
$$
Assume $x\not=y$. We can rearrange $(4)$ to get
$$
\frac{\tan\frac{1}{2}\!\!(x-y)}{\tan(x-y)}=\frac{\tan\frac{3}{2}\!\!(x+y)}{\tan(x+y)}\tag{5}
$$
Since $0<x,y\le\frac{\pi}{4}$, the left hand side of $(5)$ is positive, and the denominator of the right hand side is also positive; therefore, the numerator of the right hand side must be positive; that is, $0<\frac{3}{2}\!\!(x+y)\le\frac{\pi}{2}$. Thus, the left hand side is less than $1$ and the right hand side is greater than $1$. Contradiction; therefore, $x=y$.
