When you write down an equation in $x$ and $y$ of the form
$$F(x,y)=0,$$
then you can define a curve $\mathcal{C}$ in the plane by saying that $(x_0,y_0)\in\mathcal{C}$ if and only if $F(x_0,y_0)=0$... a point is on the curve if and only if it satisfies the equation of the curve.
Now for different $F(x,y)$ you get different curves. Some might be empty such as $x^2+y^2+1$ while others are the entire plane such as $\sin^2x+\cos^2x-1$.
Others are nicer such as $x^2+y^2-1$ which gives you a circle or the equation that gives a figure-of-eight. Now locally (i.e. near a specific point) these curves might not be that nice in that, no, they don't look the graphs of functions or differentiable functions.
For example, the point in the middle of the figure-of-eight does not look like the graph of a function. Similarly on the left- and right-hand sides of the circle there are vertical tangents --- this is not the behaviour of a differentiable function.
However away from such danger points --- such as anywhere else on the circle --- if you zoom in close enough to the point, locally, the circle looks like the graph of a function. So what we do is almost take each point $(x,y)$ on a case-by-case basis and say, O.K., near this point we could in theory define a function $y=y(x)$.
For the example of your circle. Suppose we are away from the vertical tangents and are thinking about a point $P=(x,y)$. The implicit function tells you that close to $P$ there is a function $y=y(x)$ whose graph is locally the same as the circle (in fact for the circle you have $y(x)=\pm\sqrt{1-x^2}$).
So we assume that, near $P$, we actually have (I actually get my students to write out $y=y(x)$ to help them see Chain Rules and not have $y'=1$)
$$x^2+[y(x)]^2-1=0.$$
Now these are two functions (LHS and RHS) so have the same derivative:
$$2x+2(y(x))\cdot\frac{dy}{dx}=0\Rightarrow x+y\frac{dy}{dx}=0...$$
Best Answer
Notice that if you take your equation $4x^2y^2 - 3 = 0$ then you know that
\begin{align} 4x^2y^2 &= 3\tag{1}\\ y^2 &= \frac{3}{4x^2}.\tag{2} \end{align}
Now, taking the equation $$\frac{dy}{dx} = -\frac{8xy^3}{4x^2y^2+3}$$ and using $(1)$ and $(2)$ we get $$\frac{dy}{dx} = \frac{-8xy\left(\frac{3}{4x^2}\right)}{6} = -\frac{24xy}{24x^2} = -\frac{y}{x}.$$ So, your two forms of the derivative are equivalent, given the original equation. It is frequently the case that when doing implicit differentiation you may be able to obtain several seemingly different formulations of the derivative, which are equivalent given the original equation. As we see here, sometime one is much simpler than the other.