Suppose the given fraction is equal to the integer $k$, then
$$(a+b)^3 = k(a^2+b^2)(a-b). \qquad (14)$$
First note that replacing $a$ by $da$ and $b$ by $db$ does not change the value of $k$, because numerator and denominator are both homogeneous of degree three. Therefore, we may assume $\gcd(a, b) = 1$. Moreover, when we swap $a$ and $b$, the given fraction just changes its sign. Therefore, we may assume $a > b$. Also note that $a^3 - a^2b + ab^2 - b^3 = (a^2 + b^2)(a - b)$. In particular, $a - b \mid (a + b)^3$. Because $\gcd(a - b, a + b) = \gcd(a - b, 2a) = \gcd(a - b, 2b)$ we distinguish two cases according to $a - b$ being even or odd.

Case 1: If $a - b$ is odd, $\gcd(a - b, 2b) = \gcd(a - b, b) = \gcd(a, b) = 1$, hence $\gcd(a - b, a + b) = 1$ and we can only have $a - b \mid (a + b)^3$ if $a - b = 1$. Substituting $a = b + 1$ in (14), we obtain
$$(2b + 1)^3 = k(2b^2 + 2b + 1).$$
Because $(2b + 1)^3 = 8b^3 + 12b^2 + 6b + 1 = (4b + 2)(2b^2 + 2b + 1) - (2b + 1)$, we see that $2b^2 + 2b + 1$ can only divide $(2b + 1)^3$ when it also divides $2b + 1$, but this is impossible for $b > 0$ because $2b^2 + 2b + 1 > 2b + 1$.

Case 2: If $a - b$ is even, $\gcd(a - b, 2b) = 2d$ for some positive integer $d$ which divides $b$ and $a - b$. Hence $d$ divides $\gcd(a - b, b) = \gcd(a, b) = 1$, i.e. $d = 1$. Therefore, $\gcd(a - b, a + b) = 2$ in this case. This implies that no prime $p > 2$ can divide $a - b$, since otherwise $p$ would divide $a + b$ as well according to (14). Hence, $a - b = 2^m$ for some $m \ge 1$. It is not hard to show that $m$ can only be 1, 2 or 3, but we don't need this later and therefore skip the proof.

On the other hand, $a$ and $b$ are both odd, because they are coprime and have the same parity in the current case.
When we divide $(a+b)^3$ by $a^2+b^2$, we find
$$(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = (a+3b)(a^2+b^2) - 2(a-b)b^2.$$

Together with equation (14) this implies that $a^2 + b^2$ divides $2(a - b)b^2$. If we substitute $a = b + 2^m$, where $m \ge 1$, we obtain that
$$a^2 + b^2 = 2b^2 + 2^{m+1}b + 2^{2m} \quad \text{divides} \quad 2(a-b)b^2 = 2^{m+1}b^2.$$
However, because $m \ge 1$ and $b$ is odd, the term inside the bracket
$$2b^2 + 2^{m+1}b + 2^{2m} = 2(b^2 + 2^m b + 2^{2m-1})$$
is odd, hence must divide $b^2$, which is impossible since $b^2 + 2^m b + 2^{2m-1} > b^2$. We have now established in both cases that such positive integers $a$, $b$ do not exist.