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 $x$.

It is a typo. It must be $$\tan y=\dfrac{3\sin x\cos x}{1+3\sin^2 x}=\dfrac{3\sin 2x}{5-3\cos 2x}$$for minimizing it we should have the 1st-order derivation of $\tan y$ equal to $0$ or $$\dfrac{d\tan y}{dx}=0$$which yields to $$6\cos 2x(5-3\cos 2x)=6\sin 2x\cdot 3\sin 2x$$which yields to $\cos 2x=\dfrac{3}{5}$ and $\sin 2x=\dfrac{4}{5}$. Substituting these values in the expression of $\tan y$ we attain the maximum:$$\max_{0\le x\le\dfrac{\pi}{2}} \tan y=\dfrac{3}{4}$$
Here is a sketch of $\tan y$ respect to $x$

