I think the question is self-explanatory, but I'd like to put an example of what I am referring to.
Let $C = V(f)$ be the curve defined by the polynomial $$f= -X^3 +4X^2Y-3XY^2+Y^3-2XY+Y$$ and $$F = -X_1^3+4X_1^2X_2-3X_1X_2^2+X_2^3-2X_0X_1X_2+X_0^2X_2$$ its projective completion. The point $(1,1) \sim (1:1:1)$ is singular to $C$, and I'd like to find the tangent line to that point. I think I should find the tangent cone of $f$, that is, the homogeneous terms of less degree, which in this case would be
$$V(X^2)$$ so $C$ would have a double tangent line on $(1,1)$, so the point $(1,1)$ would be a cuspid.
The thing that is bugging me off is that in almost all of the examples I've seen, the singular point is the origin $(0,0)\sim(1:0:0)$, so I'm not sure if I should just ignore the fact that the point is not the origin, or I should find the less degree terms of $f(X-1,Y-1)$ instead of $f(X,Y)$.
Furthermore, what happens when there are more than one singular point? For example, let $D=V(G)$ where $$G=X_0^2X_1^2-X_0^2X_2^2+X_1^2X_2^2$$ The points $(1:0:0), (0:1:0)$ and $(0:0:1)$ are all singular to $D$. What would be the tangent lines to those three points? Thank you in advance.
EDIT I've just realized that my second example doesn't make any sense, since that is the case where we only have one singular point $(0,0)$ on the affine space, while the other two lie in the infinity line $X_0 = 0$. I'm not sure about how to interpret that, but looking at the plot of the curve:
I guess those two points are singular because that's where the branches of the curve meet (that's roughly a wild guessing, so any help clarifying that would also be helpful).