implicit differentiation: some misconceptions If we had an equation like $xy+xy^3=2$, I know how to get $y'$.
The question is does the RHS always equals to the LHS or for certain values of $x$ only? I am so confused about this part. Does $y$ in terms of $x$ makes those two sides equal? Can we think of it as that the left part is a function that equals to the function of the right side (a function that is dependent on $y$ and $x$)?
 A: The implicit function i.e. $$x^2+xy+y^2=2\tag{1}$$
is a relation — it means it defines a set of pairs $(x,y)$. For every $x$ find such $y_1,y_2,...,y_n$ that the equation $(1)$ is true. Then our pairs are $(x, y_1),(x, y_2),...,(x, y_n)$.
For example in equation $(1)$ we put $x=1$ to get $$1+y+y^2=2$$
It has solutions $$y_1=\frac{\sqrt5-1}{2},\quad y_2=\frac{-\sqrt5-1}{2}\tag{2}$$
So at $x=1$ function $(1)$ has two values $(2)$. It turns out that such implicit functions are continuous and even differentiable, for example function $(1)$ may be plotted on WolframAlpha. If this kind of curve is differentiable it means that at some point $x$ it may have more than one derivative.
Edit: we can separate different branches from the function $(1)$. A branch is a part of some multivalued function that is single-valued. In this case we can solve for $y$ to get $$y = \frac{ \sqrt{8-3 x^2}-x}{2}\tag{a}$$
$$y = \frac{ -\sqrt{8-3 x^2}-x}{2}\tag{b}$$
Function $(a)$ is the first branch and function $(b)$ is the second. Combining plots of these two functions we get the plot of $(1)$ (try it yourself!). Having two branches of an implicit function implies that it has at most two values for one argument.
There exist other implicit or multivalued functions that may have 3 (for example complex function $\sqrt[3]{z}$), 4 or even infinitely many (i.e. complex logarithm, Lambert W-function) branches.
A: For any value of $x$, the equation defines a condition for $y$. The implicit function theorem tells us that under certain conditions, here $x\ne 0$, one can select a continuous function arc which then is as smooth as the equation.
Implicit differentiation then gives the slope of that selected function arc.
Thus you had the order of results somewhat backward, first come the function values of $y(x)$, then the derivatives.
