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Given two polynomials $f: x^3+xy+y^2$ and $g:x$ and their induced curves $C_f$ and $C_g$. I want to calculate the intersection multiplicity. So I first claim that $C_f\cap C_g=\{(0,0)\}$.

Since $f$ and $g$ have no common components I calculate the multiplicity as $\text{I}(P,C_f\cap C_g)=\sum_{P\in C_f\cap C_g} \text{I}(P,C_f\cap C_g)=\dim_k k[x,y]/(f,g)=\dim_k k[x,y]/(x,y)=1$.

But how does this fits to Bezout's theorem which states $$\sum\text{I}(P,F\cap G)=\deg F\cdot\deg G$$ for projectives curves $F,G$?

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    $C_f\cap C_g=\{(0,0),(-2,-2)\}$. – 2017-02-24
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    But $g(x,y)=x$ has no zero in $(-2,-2)$ – 2017-02-24
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    Sorry, I misread the equation of $g:$: I thought of the line $y=x$. I'll remove my comment in a moment. Please see my answer. – 2017-02-24

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$$k[x,y]/(f,g)=k[x,y]/(x^3+xy+y^2,x)\simeq k[y]/(y^2),$$ so you find a value of $2$, but you forget the equality is valid for projective curves. For affine curves, $\deg f\deg g$ is only an upper bound.

Consider the associated projective curves: $X^3+XYT+Y^2T$ and $X$, and the graded ring $$k[X,Y,T]/(X^3+XYT+Y^2T,X)\simeq K[Y,T]/(Y^2T),$$ which is an artinian ring of length $3$. On the other hand, $(0:0:1)$ has multiplicity $2$, and the point at infinity $(0:1:0)$ has multiplicity $1$.

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    But $K[Y,T]/(Y^2T)$ hat dimension...? It might be dimension $3$ since Bezout must hold, mustn't it? – 2017-02-24
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    If you mean the Krull dimension, you have to consider the associated affine variety $k[y]/(y^2)$, which has dimension $0$. However, the sum of the intersection multiplicities is the length of the ring, which is $3$. – 2017-02-24