I'm having issues cranking through this problem. I think I'm not understanding the question at a more fundamental level, so any help is appreciated:
Find $dy/dx=y'(x)$ if $x$, $y$ are related by $\sin(x)+\cos(y)=\sin(x)\cos(y)$.
Here are my thoughts: I can divide by a $\sin(x)$ to get $1+\cot(y/x)=\cos(y)$, thus giving me $\cos(y)-\cot(y/x)=1$, and then go from there? This is a confusing problem for me. Any help is great. Thanks!
The first thing to realize is that this is not a "multivariable" calculus problem. While there are clearly two variables, you want to think of $y$ as a function of $x$, which is a standard single-variable situation.
The technique used to solve problems like these, where x and y are related with an equation, but not in the form of $y = f(x)$, is called implicit differentiation. The trick is to realize that, as you stated at first, $y$ is really a function of $x$ (you called it $y(x)$), not an independent variable.
So you want to differentiate both sides of the equation with respect to $x$, and when you see $y$, think of it as a function of $x$. This means that the (single-variable) chain rule must be used. For example, $\dfrac{d}{dx}(\cos y) = \dfrac{d}{dx}(\cos (y(x)) = -\sin y \cdot y'(x)$.
In this problem we have:
$\frac{d}{dx}(\sin x + \cos y) = \frac{d}{dx}(\sin x \cos y)$
$\cos x -\sin y \cdot y'(x) = \cos x \cos y - \sin x \sin y \cdot y'(x)$.
This gives you an equation that you can solve for $y'(x)$. Note that the other side will involve both $x$ and $y$, which is pretty common. You usually don't need to solve for $y$ in terms of $x$ or anything.