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Answering your questions a bit out of order, I'll start with the "non-rigorousness" of talking about $dx$ by itself. While this seems to be commonly poorly explained, the derivative (of a function from and to reals) is an operation that takes functions to functions. Let's consider the typical high school definition of the derivative: $$\frac{df}{dx}(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$

This takes a function $f$ and a number $x$, and gives you a new number $\lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$. There is no $dx$ piece you can take off. To this end, Newton's $\dot f$ notation is more honest and coherent. To rationalize, "setting $dx=0$", you could have just as well said you were setting $\frac{dx}{dy} = 0$. Indeed, that's exactly what you do in the following part. "Intuitively", assuming $b\neq 0$ so $a/b$ is well-defined, then $a/b = 0$ if and only if $a = 0$. In Leibniz' notation, we have $dx/dy = \frac{1}{dy/dx}$ or $\frac{dy}{dx}\frac{dx}{dy} = 1$. This is actually hiding a fairly important and non-trivial result known as the [inverse function theorem](https://en.wikipedia.org/wiki/Inverse_function_theorem). Informally, the inverse function theorem states that if a function has a non-zero derivative in some sufficiently small open interval around some point, then that function has an inverse on that interval (whose derivative is necessarily the reciprocal of the original function). This is why we can talk about $x$ as a function of $y$ and take its derivative.

Ultimately, I think Leibniz notation is extremely misleading and that it doesn't scale well to higher dimensions. There's more I could say here, but I won't.

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For your other questions, the first question you should ask yourself is "what is the definition of a tangent line?"

Intuitively speaking, there are (at least) two obvious ways to think about this for a curve.

One approach would be to parameterize the curve, i.e. think of it as a trajectory in "time". Given a well-behaved parameterization, our "velocity" along the trajectory, i.e. the derivative of the trajectory, will be a vector tangent to the curve. One property we want for the trajectory to be well-behaved is that its derivative is never zero. Thus, this gives us a non-zero tangent vector, and thus tangent line, at every point of the *parameter space*. If two points in the parameter space, i.e. in "time", map to the same point on the curve, then we could potentially have two (or more) tangent vectors at that point of the curve. Indeed, in the self-intersecting case you provide, this is what would happen.

The big downside of this approach is that we need a suitable parameterization. So another intuitive approach is to view the curve as a [level curve](https://en.wikipedia.org/wiki/Level_set) or contour of a real function on the plane, $\varphi$. In your case, $\varphi(x,y) = x^2 + y^3 - 15xy$ and the curve is the level curve determined by $\varphi(x,y) = 0$. Intuitively, a tangent vector at a point would be a direction that keeps us on the curve, i.e. a direction in which $\varphi$ does not change. More formally, we could say it is the vectors $\mathbf v$ such that the directional derivative of $\varphi$ in the direction $\mathbf v$ (at a given point $(x,y)$) is $0$. Using the vector derivative, we can write the "derivative in the $\mathbf v$ direction" operator as $\mathbf v \cdot \nabla$, and thus we're asking for $\mathbf v \cdot \nabla\varphi = 0$. $\nabla \varphi$ is also the [gradient](https://en.wikipedia.org/wiki/Gradient) of $\varphi$, and, in this case, we have that $(\mathbf v \cdot \nabla)\varphi = \mathbf v \cdot (\nabla \varphi)$. $\mathbf v \cdot (\nabla \varphi) = 0$ is the statement that $\mathbf v$ is in the [orthogonal complement](https://en.wikipedia.org/wiki/Orthogonal_complement#Inner_product_spaces) of the gradient of $\varphi$. The orthogonal complement to a vector in 2D is a line.

In this case, it is indeed the tangent line to the point on the curve. Intuitively, the gradient tells us the direction of greatest increase and its negative would be the direction of greatest decrease. We want the direction that makes no change. If we move in a direction with a non-zero projection onto the gradient, then we will increase or decrease the value of $\varphi$ and move off the curve. Thus any tangent vector must have zero projection onto the gradient, which is another way of saying it's in the orthogonal complement.

In some sense, what you're doing is this latter approach albeit you're hamstrung by the need to have $y$ be a function of $x$ or vice versa which adds complexity. The gradient of $\varphi$ is $(2x - 15y, 3y^2 - 15x)$. The vertical tangent lines will occur where the second component is zero (remember we want the *orthogonal complement* to this vector), and the horizontal tangent lines will occur where the first component is zero. Of course, there's nothing special about vertical and horizontal. We can compute the tangent line at any point on the curve by computing the orthogonal complement of this vector at that point. This is supposing that the gradient isn't zero which is what you're running into at the self-intersection at $(0,0)$.

It's questionable whether you'd want to consider *any* line a tangent line at a self-intersection like this, which is part of why I said you should think about what your definition of "tangent line" even is. Nevertheless, in this case one solution would be to take the limit as we approach this point. This doesn't help by itself as the limit is still zero. Continuity requires that the magnitude continuously decreases to zero to get a zero vector. But we only care about the direction, so we could instead take the limit of the *normalized* gradient, i.e. $\nabla\varphi / |\nabla \varphi|$. This leads to a *discontinuity* at $(0,0)$. In this case, this means the limit we get depends on how we approach it. For your purposes, you can then just take the limit with the $x$-component constant to test for vertical tangent lines, and again with the $y$-component constant to test for horizontal tangent lines. For this example, taking the limit with $x=0$ as $y$ approaches $0$ will lead to a normalized gradient with a non-zero $x$-component and zero $y$-component, and thus a vertical tangent. If we take the limit with $y=0$ as $x$ approaches $0$, we end up with a normalized gradient where neither component is zero, so neither a horizontal nor vertical tangent line.