algebraic geometry

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...]