Tangents at singularities Given an implicit polynomial function $f$ with singularity at the origin. How do I find the tangents to the curve at the point?
Wikipedia says that ignore all the terms except the ones with lowest degree. Why is this true?
Take for example the curve $x^3+x^4+y^3=0$. What are the tangents at the origin?

Wikipedia is right: the tangent cone to your curve $C$ is given by $x^3+y^3=0$.
If the base field is algebraically closed of characteristic $\neq 3$ and if $j\neq 1$ is a primitive cubic root of$1$, it consists of the three lines $x+y=0, x+jy=0, x+j^2y=0$, since $x^3+y^3=(x+y)(x+jy)(x+j^2y)$.  
But where does the Wikipedia recipe come from? 
Here is an explanation: we are looking at the intersection at $(0,0)$ of our curve with the line $L_{ab}$ given parametrically by  $x=at, y=bt$ (with $a,b$ not both zero).
The values of $t$ corresponding to an intersection point $L_{ab}\cap C$ are those satisfying the equation $(at)^3+(at)^4+(bt)^3=0$ gotten by substituting $x=at, y=bt$ into the equation of $C$.
The equation for $t$ is thus $$t^3(a^3+at+b^3)=0 \quad (MULT)$$ The result is that $t=0$ is a root of multiplicity $3$ for all values of $a,b$ except for those with $a^3+b^3=0$ for which the multiplicity of the zero root of $(MULT)$ is $4$.
So we decree that the lines $L_{ab}$ with $a^3+b^3=0$ are the tangents to $C$ at the origin because they cut $C$ with a higher multiplicity, namely $4$, than all the other lines which only cut $C$ with multiplicity $3$.
This calculation is easily generalized to arbitrary curves through the origin and justifies Wikipedia's recipe.  
[Purists will notice that the above is purely algebraic: no limits are involved and we don't need a topology on the base field.
 But assuming that the base field is $\mathbb C$ with its metric topology and using limits is fine with me if it helps in understanding the situation. In mathematics as in love and war all is fair...]
