6. From the set of positive integers not exceeding 2013, $\{1,2, \cdots, 2013\}$, any two positive integers are chosen. Then the probability that they are exactly the solutions to the equation $x^{3}+y^{3}=x^{2} y^{2}$ is
6. $\frac{1}{2013^{2}}$.

First, study the positive integer solutions of $x^{3}+y^{3}=x^{2} y^{2}$.
Let $(x, y)$ be a solution of the equation satisfying $x \geqslant y$. Then
$$
\begin{array}{l}
x^{2} \mid y^{3} \Rightarrow y^{3} \geqslant x^{2} \\
\Rightarrow 4 y^{3} \geqslant 4 x^{2} .
\end{array}
$$

Since $x \geqslant y$, we have $x^{2} y^{2}=x^{3}+y^{3} \leqslant 2 x^{3}$.
Thus, $y^{2} \leqslant 2 x$.
Hence, $y^{4} \leqslant 4 x^{2}$.
From (1) and (2), we get $y \leqslant 4$.
When $y=4$, the rational solutions of $x^{3}-16 x^{2}+64=0$ can only be the integer divisors of 64, and none of them satisfy the equation upon inspection;

When $y=3$, the rational solutions of $x^{3}-9 x^{2}+27=0$ can only be the integer divisors of 27, and none of them satisfy the equation upon inspection;

When $y=2$, the rational solutions of $x^{3}-4 x^{2}+8=0$ can only be the integer divisors of 8, and upon inspection, only $x=2$ satisfies;

When $y=1$, the rational solutions of $x^{3}-x^{2}+1=0$ can only be the integer divisors of 1, and none of them satisfy the equation upon inspection.
Therefore, the equation has only one solution $(x, y)=(2,2)$.
Thus, the required probability $p=\frac{1}{2013^{2}}$.