I'm currently brushing up on my calculus and was studying about implicit differentiation. The method proposed in the textbook I'm using (University Calculus: Alternate Edition (Hass et al., 2008) for anyone curious) and the method that the authors instruct is that if there are two variables $x$ and $y$, then differentiate both sides of the equation and treat $y$ as a function of $x$.
What I'm confused about is that there are many exercises that instruct to find the tangent or normal line at a certain point using implicit differentiation. This operation isn't hard itself, but what I'm confused is whether or not it's correct to treat $y'$ as a function of both $x$ and $y$. For example:
Find (a) the tangent and (b) the normal to the curve at the given point $P(-1, 0)$.
$$6x^2 + 3xy + 2y^2 + 17y - 6 = 0$$
Differentiating both sides w.r.t. $x$ gives us:
$$12x + 3y + 3xy' + 4yy' + 17y' = 0$$
and finally:
$$y' = -\dfrac{12x + y}{3x + 4y + 17}$$
Finding the slope of the tangent line means plugging in $x=-1$ and $y=0$. Is it correct to write:
$$y'(-1, 0) = -\dfrac{12\cdot(-1) + 0}{3\cdot(-1) + 4\cdot 0 + 17} = \frac{6}{7}$$
What you have done is correct. Conceptually, you already have $y$ being a function of $x$. Here you can find it explicitly by putting your original equation into the quadratic formula. You could then differentiate that and get $y'$ as an explicit function of $x$. This means $y'$ is just a function of $x$. The expression you have uses $y$, which is a function of $x$, because it is convenient.