A circle $x^2 + y^2 = r^2$, with $x$ and $y$ the coordinates of a point on it and $r$ its radius, is a relation, not a function written $y = \ldots$; solving for $y$ gives a root and a choice of sign. Implicit differentiation skips the solving: treat $y$ as a function of $x$ that happens not to be written out, and differentiate both sides with respect to $x$. The $x^2$ term gives $2x$. The $y^2$ term is a function of $x$ SQUARED, so the chain rule applies, with $u = y$: it gives $2y \cdot \dfrac{dy}{dx}$, and that $\dfrac{dy}{dx}$ is the trap in the lesson's name. Writing $2y$ alone, with no $\dfrac{dy}{dx}$, leaves nothing to solve for. The constant $r^2$ gives 0.

So $2x + 2y\dfrac{dy}{dx} = 0$, and $\dfrac{dy}{dx} = -\dfrac{x}{y}$, read aloud: dee y by dee x equals minus x over y. At $(3, 4)$ on $x^2 + y^2 = 25$ the slope is $-\dfrac{3}{4} = -0.75$. Check it: the radius to $(3, 4)$ has slope $\dfrac{4}{3}$, the tangent is perpendicular to the radius, and $\dfrac{4}{3} \cdot \left(-\dfrac{3}{4}\right) = -1$. The slope depends on the point, both coordinates, and not on $r$; and where $y = 0$ the tangent is vertical, which is where $-\dfrac{x}{y}$ divides by zero.