Consider these equations:
$$ \begin{gather} x^3 + y^3 = 6xy \\ x^2 + 3y^2 + 4y = 4 \end{gather}$$
These equations implicitly define \(y\) as a function of \(x\). They can be written as follows:
$$ \begin{gather} x^3 + y^3 - 6xy = 0 \\ x^2 + 3y^2 + 4y - 4 = 0 \end{gather}$$
We can define a function using the left hand side:
$$ \begin{gather} H(x,y) = x^3 + y^3 - 6xy \\ G(x,y) = x^2 + 3y^2 + 4y - 4 \end{gather}$$
In the first case, \(H(x,y)=0\) defines \(y\) as an implicit function of \(x\), and in the second case, \(G(x,y)=0\) defines \(y\) as an implicit function of \(x\).
If an equation of the form \(F(x,y)=0\) defines \(y\) as an implicit differentiable function of \(x\), and \(F\) is differentiable, we can use the chain rule to state:
$$\frac{∂F}{∂x} \frac{dx}{dx} + \frac{∂F}{∂y} \frac{dy}{dx} = 0 $$
In other words:
$$ \frac{dy}{dx} = -\frac{\frac{∂F}{∂x}}{\frac{∂F}{∂y}}$$