How can I show that $y\sin(x+y) = 0$ defines $x$ implicitly as a function of $y$ in a neighbourhood of $(0, \pi)$ and calculate its derivative $\dfrac{dx}{dy}\Bigg|_{y=\pi} ?$
I'm struggling with the implicit theorem for multivariable calculus. My attempt was:
$$\frac{d}{dy}(y\sin(x+y)) = 0$$ $$\iff \sin(x+y)\frac{d}{dy}y + y\frac{d}{dy}\sin(x+y) = 0$$ $$\iff \sin(x+y) + y\cos(x+y)\frac{d}{dy}(x+y) = 0$$ $$\iff \sin(x+y) + y\cos(x+y)\frac{d}{dy}x + y\cos(x+y) = 0$$ $$\iff \frac{dx}{dy} = -\frac{\sin(x+y) + y\cos(x+y)}{y\cos(x+y)}$$
But it doesn't seem right and neither I can calculate $\dfrac{dx}{dy}\Bigg|_{y=\pi}$ because it is still dependent on $x$. What am I doing wrong and how should I think about it?
Since you need to show 'implicit'-ness in a neighborhood of $(0,\pi)$, this is a hint that you need to substitute $(x,y)=(0,\pi)$ in your final expression for $\dfrac{dx}{dy}$. Your calculated expression for $\dfrac{dx}{dy}$ is correct. Substituting, we have $$\dfrac{dx}{dy}\Bigg|_{(x,y)=(0,\pi)} = -\frac{\sin(0+\pi) + \pi\cos(0+\pi)}{\pi\cos(0+\pi)} = -\frac{0-\pi}{-\pi}=-1.$$ This is the answer. Now, substituting $(x,y)=(0,\pi)$ works because your are in a neighborhood of this point in the plane.