<s>[INST] Given $x, y$ are acute angles such that $\sin y= 3\cos(x+y)\sin x$. 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 at a solution:
$$(\sin y = 3(\cos x \cos y - \sin x \sin y) \sin x) \frac{1}{\cos y}$$
$$\tan y = (3 \cos x - 3 \sin x \tan y) \sin x$$
$$\tan y + 3 \sin^2 x + \tan y = 3 \sin x \cos x$$
$$\tan y = \frac{3 \sin x \cos x}{1 + 3 \sin^2 x}$$
I have also tried substituting $0$, $30$, $45$, $60$, $90$ to the values of $x$.
 [/INST] Starting from marty cohen's answer
$$\tan (y)=\dfrac{3\sin(2 x)}{5-3\cos(2x)}$$ let $t=\tan(x)$ to get 
$$\tan (y)=\frac{3 t}{4 t^2+1}=f(t)$$ So 
$$f'(t)=\frac{3-12 t^2}{\left(4 t^2+1\right)^2}\qquad \text{and} \qquad f''(t)=\frac{24 t \left(4 t^2-3\right)}{\left(4 t^2+1\right)^3}$$ The first derivative cancels for $t_{\pm}=\pm \frac 12$. For $t_+$, $f''(t+)=-3$ corresponding to a maximum and for $t_-$,  $f''(t_-)=3$ corresponding to a minimum. 
So, $t=\frac 12$ and $\tan(y)=\frac34$
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