Introduction

When we have $y$ explicitly defined as a function of $x$, say $y=f(x)=x^2$,  we can find $\frac{dy}{dx}$ by differentiating $x^2$.  In other circumstances, we know $y=f(x)$ is a function of $x$, but we do not know what $f$ is, so we say that $y$ is implicitly defined as a function of $x$.  In this situation, to differentiate $y$, we get $\frac{d}{dx}(y)=\frac{dy}{dx}$ (of course, these expressions are the same), because we do not know what function to differentiate.  

Why we care

In addition to taking derivatives of functions we can take derivatives of equations. After all, if both sides of an equation are always equal, then their rates of change are also equal.  If the equation involves both $x$ and $y$, and we are differentiating with respect to $x$, then $\frac{d}{dx}(y)=\frac{dy}{dx}$ as explained above, because that is the best we can do with the derivative of $y$ without knowing the formula for $y$. 

Notice that we have curves for which $y$ is not explicitly a function of $x$, such as the unit circle $x^2+y^2=1$.  Here, if we solve for $y$, we get $y=\pm\sqrt{1-x^2}$ which is not a function.  (Why not? Think about this before reading more.  To see why:  Plug in a value for $x$, say $x=0$, and you get two values for $y$, namely $1$ and $-1$.  You can also see this curve does not represent a function by looking at the vertical line test.)  However, we can still find $\frac{dy}{dx}$, by using implicit differentiation, by using the values of $x$ and $y$, not just $x$, to decide where we are differentiating.  This allows us to use the valid function, either $y=\sqrt{1-x^2}$ or $y=-\sqrt{1-x^2}$ depending on whether $y$ is positive or negative.