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? 

$$\tan y=\dfrac{3\tan x}{1+4\tan^2x}$$
$$\iff(4\tan y)\tan ^2x-3\tan x+\tan y=0$$ which is a Quadratic Equation $\tan x$
As $\tan x$ is real, the discriminant must be $\ge0$
i.e., $$3^2-4(4\tan y)\ge0\tan y\iff\tan^2y\le\dfrac9{16}\iff-\dfrac34\le\tan y\le\dfrac34$$
