$$x\cos (4x+3y)=y\sin x$$
I have been stuck on this problem for the longest. I have the answer but I don't know how to get to it. I have used the product and chain rule on both sides. I keep getting this:
$$\frac{\cos(4x+3y)-4x\sin(4x+3y)-y\cos x}{\sin x}$$
Here is the answer:
$$\frac{\mathrm dy}{\mathrm dx} = \frac{\cos (4x+3y)-y\cos x-4x\sin (4x+3y)}{\sin x + 3x\sin (4x+3y)}$$
$\sin(4x+3y)+x\cos(4x+3y)(4+3dy/dx)=(dy/dx)\cos x -y\sin x$
$\sin(4x+3y)+4x\cos(4x+3y)+3(dy/dx)\cos(4x+3y)=(dy/dx)\cos x -y\sin x$
$\sin(4x+3y)+4x\cos(4x+3y)+y\sin x=(dy/dx)\cos x + 3(dy/dx)\cos(4x+3y)$
$\sin(4x+3y)+4x\cos(4x+3y)+y\sin x=(dy/dx)(\cos x + 3\cos(4x+3y))$