<s>[INST] If $x$ and $y$ are acute, and $\sin y = 3 \cos (x+y) \sin x$⁡, then find the maximum value of $\tan y$ 
Given $x,y$ are acute angles such that
  $$\sin y = 3 \cos(x+y)\sin x$$
  Find the maximum value of $\tan ⁡y$.

Attempt:
We have
$$\begin{aligned} 3(\cos x \cos y - \sin x \sin y) \sin x & = \sin y \\ 3 \cos x \sin x - 3 \sin^2 x \tan y & = \tan y \\ 3 \cos x \sin x & = \tan y(1 +3 \sin^2 x) \\ \tan y & = \dfrac{3 \sin x \cos x} {1+3 \sin^2 x} \end{aligned}$$
Now, how about the next step? Or maybe I did some mistakes? 
 [/INST] Put $t = \tan x$, then $\tan y = \dfrac{3t}{1+4t^2}\le \dfrac{3t}{4t} = \dfrac{3}{4}$, which is the max of $\tan y$ with equality occurs when $t = \dfrac{1}{2}$ or $2\sin x = \cos x$...the rest is simple...
</s>