Answer this one. The old one is wrong.

$\frac{1}{x}+\frac{1}{y}=\frac{1}{3}$

$\frac{1}{y}=\frac{1}{3}-\frac{1}{x}$

$\frac{1}{y}=\frac{x-3}{3x}$

$y=\frac{3x}{x-3}=\frac{(3x-9)+9}{x-3}$

$y=3+\frac{9}{x-3}$

Since $y$ is a positive integer, then $x-3$ must divide $9$ . So, the possible values of $x$ are $2,3,4,6$ and $12$ . No real numbers can be divided by zero, so we eliminate $3$ .

From here, ordered pairs satisfying the conditions above are $(2,-6),(4,12),(6,6),(12,4)$ . But $y$ cannot be negative, and we are left with $\boxed{3}$ ordered pairs.