A relation between $x$ and $y$ does not always solve explicitly for $y$. If the relation determines $y$ locally as a function of $x$, we can still differentiate it by treating $y$ as $y(x)$ and applying the chain rule.

For example, from

$$x^2+y^2=25$$

we get

$$2x+2y\frac{dy}{dx}=0$$

so wherever $y\ne 0$,

$$\frac{dy}{dx}=-\frac{x}{y}$$

The extra factor $dy/dx$ appears because differentiating a term involving $y(x)$ requires the chain rule. Implicit differentiation therefore extracts local rates of change directly from a constraint, without first isolating one variable.

Visual intuition: implicit differentiation