6. (40 points) Let $f(x)$ be a polynomial of degree 2010 such that $f(k)=-\frac{2}{k}$ for $k=1,2, \cdots, 2011$.
Find $f(2012)$.
6. Let $g(x)=x f(x)+2$.

Since the degree of $f(x)$ is 2010, the degree of $g(x)$ is 2011.
$$
\text { Also, } g(k)=k f(k)+2=0 \text { for } k=1,2, \cdots, 2011, \text { then }
$$
$$
g(x)=\lambda(x-1)(x-2) \cdots(x-2011),
$$

where $\lambda$ is some real number.
Let $x=0$. Then $2=-2011! \cdot \lambda$.
Thus, $g(2012)=2011! \cdot \lambda=-2$.
Therefore, $f(2012)=-\frac{1}{503}$.