From my knowledge so far, pretty much all functions take an input and spit out some output. However, with implicit functions, this isn't necessarily the case, and that intuition breaks apart for me. For instance, in:
$$y + \ln y = \sin x$$
My intuition goes out the window. With input $x$, we get.. output $y$ summed with the natural logarithm of output $y$. Does anyone have any explanation on how to properly comprehend this?
Yes, the nature of an implicit function is to obscure this input-output relationship. In fact, generally if you write something of the form $g(x,y) = 0,$ it won't be a function at all, but a relation.
Nonetheless if the definition of a function is satisfied (i.e. for every $x$ in some domain there is a unique $y$ satisfying $g(x,y) = 0$), your implicit function is sure-as-day a function and there should really be no inhibition to thinking of it as an input-output relationship. After all, every $x$ has a unique $y$ associated with it. So $x$ is the input and $y$ is its output.
Your example $y+\ln(y) = \sin(x)$ is in fact a true function. The way you can tell this easily is that the left-hand side is an increasing, and thus one-to-one, function of $y$ on its domain $(0,\infty).$
You can even use this fact to make the input-output relationship clearer. Let $h(y) = y+\ln(y).$ As I said before, this is an increasing, thus one-to-one, thus invertible function. So you can write your relation as $$ h(y) = \sin(x)$$ and then $$ y = h^{-1}(\sin(x))$$ where $h$ is the inverse that we know exists thanks to the increasingness of $h.$ Hey! We've written the function in an explicit form (even though we've had to use an unfamiliar function--the inverse of $y+\ln(y)$-- to do so).
But don't let the fact that this one has a nice way to write it in explicit form allow you forget the what I said at the end of the second paragraph. Regardless if you can get it into explicit form or not, if a relation happens a function, you can think of it as input-output... that goes hand in hand with the definition of function.