A student adds up rational fractions incorrectly:

\[\frac{a}{b}+\frac{x}{y}=\frac{a+x}{b+y}\quad (\star) \]

Despite that, he sometimes obtains correct results. For a given fraction $\frac{a}{b},a,b\in\mathbb{Z},b>0$, find all fractions $\frac{x}{y},x,y\in\mathbb{Z},y>0$ such that the result obtained by $(\star)$ is correct.
To solve the problem, we need to find all fractions \(\frac{x}{y}\) such that the incorrect addition formula 

\[
\frac{a}{b} + \frac{x}{y} = \frac{a+x}{b+y}
\]

yields a correct result. This means we need to find \(\frac{x}{y}\) such that:

\[
\frac{a}{b} + \frac{x}{y} = \frac{a+x}{b+y}
\]

is equivalent to the correct addition of fractions:

\[
\frac{a}{b} + \frac{x}{y} = \frac{ay + bx}{by}
\]

We start by equating the two expressions:

\[
\frac{a+x}{b+y} = \frac{ay + bx}{by}
\]

Cross-multiplying to clear the denominators, we get:

\[
(a + x)by = (ay + bx)(b + y)
\]

Expanding both sides, we obtain:

\[
aby + bxy = aby + ay^2 + b^2x + bxy
\]

Subtracting \(aby + bxy\) from both sides, we get:

\[
0 = ay^2 + b^2x
\]

Rearranging the terms, we find:

\[
ay^2 = -b^2x
\]

Solving for \(x\), we get:

\[
x = -\frac{a}{b^2} y^2
\]

Since \(x\) and \(y\) must be integers, we need \(\frac{a}{b^2}\) to be an integer. Let \(k\) be an integer such that:

\[
y = bk
\]

Substituting \(y = bk\) into the equation for \(x\), we get:

\[
x = -\frac{a}{b^2} (bk)^2 = -\frac{a}{b^2} b^2 k^2 = -a k^2
\]

Thus, the pair \((x, y)\) that satisfies the condition is:

\[
(x, y) = (-a k^2, b k)
\]

where \(k\) is any integer.

The final answer is \( \boxed{ (x, y) = (-a k^2, b k) } \) where \(k \in \mathbb{Z}\).