Second derivative of $x \sin ⁡y + y \sin x = \pi$ WRT $x$ Can anyone help me with this? I can't find the second derivative of
$$x \sin ⁡y + y \sin x = \pi$$

edit: below are my working... I feel like I'm going no where...
My working:
 A: $$ x\sin y+y \sin x=\pi$$
$$\sin y +\frac{dy}{dx}x\cos y+y\cos x+\frac{dy}{dx}\sin x=0$$
For $(\frac{\pi}{2},\frac{\pi}{2})$
$$\frac{dy}{dx}=-1$$
$$\frac{dy}{dx}\cos y+\frac{dy}{dx}\cos y+\frac{d^2y}{dx^2}x\cos y-(\frac{dy}{dx})^2x\sin y+\frac{d^2y}{dx^2}\sin x+\frac{dy}{dx}\cos x=0 $$
For $(\frac{\pi}{2},\frac{\pi}{2})$
$$\frac{d^2y}{dx^2}=\frac{\pi}{2}$$
Try not to make it in quotient form. Quotient rule is very difficult and complicated. Try to avoid it if can. 
$$\frac{d}{dx}\sin y=\frac{dy}{dx}\cos y$$
By chain rule
$$\frac{d}{dx}F(g(x))=g'(x)F'(g(x))$$
$$\frac{d}{dx}(\frac{dy}{dx}x\cos y)=\frac{d(x)}{dx}(\frac{dy}{dx}\cos y)+\frac{d(\frac{dy}{dx})}{dx}x\cos y+\frac{d(\cos y)}{dx}\frac{dy}{dx}x$$
$$\frac{dy}{dx}\cos y+\frac{d^2y}{dx^2}x\cos y-(\frac{dy}{dx})^2x\sin y$$
Know that $$\frac{dy}{dx}\frac{dy}{dx}=(\frac{dy}{dx})^2$$
$$\frac{d(\frac{dy}{dx})}{dx}=\frac{d^2y}{dx^2}$$
A: For the first derivative we have
$$\sin(y)+x\cos(y)y'+y'\sin(x)+y\cos(x)=0$$
solving for $y'$ we obtain
$$y'=-\frac{\sin(y)+y\cos(x)}{\sin(x)+x\cos(y)}$$ and for the second derivative we get
$$\cos(y)y'+\cos(y)y'+x(-\sin(y))y'^2+y''x\cos(y)+y''\sin(x)+y'\cos(y)-y\sin(x)=0$$
A: Use product rule and chain rule.
$(x\sin y+y\sin x)=0\implies (x\sin y+y\sin x)'=\pi'\implies (\sin y+\cos y\cdot \frac{dy}{dx}+\sin x\cdot\frac{dy}{dx}+\cos x\cdot y)=0$.
Now continue....
