Answer
$$\frac{dy}{dx}=\frac{\cos(2x+3y)-2x\sin(2x+3y)-y\cos x}{\sin x+3x\sin(2x+3y)}$$
Work Step by Step
$$x\cos(2x+3y)=y\sin x$$
1) Differentiate both sides of the equation with respect to $x$:
$$\frac{d}{dx}(x\cos(2x+3y))=\frac{d}{dx}(y\sin x)$$
$$\cos(2x+3y)+x\frac{d}{dx}(\cos(2x+3y))=\frac{dy}{dx}\sin x+y\cos x$$
$$\cos(2x+3y)+x(-\sin(2x+3y))\frac{d}{dx}(2x+3y)=\frac{dy}{dx}\sin x+y\cos x$$
$$\cos(2x+3y)-x\sin(2x+3y)(2+3\frac{dy}{dx})=\frac{dy}{dx}\sin x+y\cos x$$
$$\cos(2x+3y)-2x\sin(2x+3y)-3x\sin(2x+3y)\frac{dy}{dx}=\frac{dy}{dx}\sin x+y\cos x$$
2) Collect all the terms with $dy/dx$ onto one side and solve for $dy/dx$:
$$\frac{dy}{dx}\sin x+3x\sin(2x+3y)\frac{dy}{dx}=\cos(2x+3y)-2x\sin(2x+3y)-y\cos x$$
$$\frac{dy}{dx}=\frac{\cos(2x+3y)-2x\sin(2x+3y)-y\cos x}{\sin x+3x\sin(2x+3y)}$$
