Intersection multiplicity for two curves defined by $f=0,g=0$ I want to understand how I can find the intersection multiplicity $I_p$ at a point $p$ for two curves $f,g$. 

I have the example where 
  $$ f(x,y) = y^2-x^3, \,\,\,\, g(x,y)=y^2-x^2(x+1) $$
  Then I am looking for the intersection multiplicity of the ideal $(f,g)$ at $p=(0,0)$.

My first attempt is to notice that $g(x,y)=y^2-x^3-x$ and the ideal $(f,g)$ can be written as $(f,g-f)=(y^2-x^3,x^2)$. But now I am stuck since I cannot find a way to possibly simplify this ideal as to conclude about the intersection form. Only step I can go further is to note that
$$ I_{(0,0)}(y^2-x^3,x^2) = 2I_{(0,0)}(y^2-x^3,x) $$
What would the next step be in order to determine $I_p$? In a previous example I was able to reduce the original ideal to an ideal involving only degree 1 curves concluding easily about what the multiplicity is. Now? 
 A: It doesn't look like anyone ever responded to your question, so here's an answer many days late (and many dollars short).
As Mohan said, $(y^2 - x^3, x^2) = (y^2, x^2)$.  Note that since $x^2 \in (y^2 - x^3, x^2)$, then $x^3 = x \cdot x^2 \in (y^2 - x^3, x^2)$.  Then $y^2 = y^2 - x^3 + x^3 \in (y^2 - x^3, x^2)$ as it a sum of elements of $(y^2 - x^3, x^2)$.  This shows $(y^2 - x^3, x^2) \supseteq (y^2, x^2)$, but in fact the reverse inclusion holds as well since $y^2 - x^3 = y^2 - x \cdot x^2 \in (y^2, x^2)$.
Fulton lists 7 axioms that you can use in computing intersection multiplicity in section 3.3 (pp. 36-37) of his book Algebraic Curves.  (The one I used above is axiom 7.)
Now we have to find the dimension over $k$ of the local ring $\left(\frac{k[x,y]}{(y^2,x^2)}\right)_{(x,y)}$.  One can show that you don't actually need to localize, and $1, x, y, xy$ forms a basis for the quotient ring, so the answer is 4, as stated in the comments.  (This also follows from axiom 6 of Fulton's book.)
