It is well known that, given a function $f:\mathbb{R} \to \mathbb{R} $, $f'(x_0)$ can be interpreted as the slope of the tangent line to $f$ in $x_0$. What about curves of the form $F(x, y, c)=0$, which cannot be written in the form $y=f(x)$, like the circumference $x^2+y^2-c^2=0$? I know that in this example, to find the tangent in the point $(x_0, y_0)$, one may distinguish the two functions $y=\sqrt{c^2-x^2}$ and $y=-\sqrt{c^2-x^2}$ and then evaluate the derivative in $x_0$(now ignoring $y_0$, since those are functions), but this procedure doesn't really convince me, as in general one should actually know how to "split up" the curve into two (or, I guess, more) functions. Thus, I have asked my Professor for a general working rule and he answered to use implicit differentiation. For example, suppose we want to find the slope of the tangent to $x^2+y^2-25=0$ in the point $(3,-4)$. Then, differentiating both sides we get $$2x+2y \frac{dy}{dx}=0 \implies \frac{dy}{dx}=-\frac{x}{y}$$ Hence, the slope in $(3,-4)$ is simply $-\frac{3}{(-4)} = \frac{3}{4}$ . This is indeed the correct answer (in fact, the same I would get if I computed the derivative of $-\sqrt{25-x^2}$ and evaluated it in $x=3$). However, I am very confused about the meaning of $\frac{dy}{dx}$ here, since again this is not a function, and I do not understand the geometric interpretation of implicit differentiation either (that is, how/why this trick works).
In the same way, my Professor showed that for every $(a,b) \in \mathbb{R^2}$ the curves $$x^2-y^2=a$$ and $$xy=b$$ are orthogonal, since $$x^2-y^2-a=0 \implies \frac{dy}{dx}=\frac{x}{y}$$ and since $$xy-b=0 \implies \frac{dy}{dx}=-\frac{y}{x}$$Could you please clarify what I have asked before? Thanks in advance for your precious time and kindness.