Let $I = a - \frac{1}{b} + b(b + \frac{3}{a})$. If we want $I = a + b^2 + \frac{3b}{a} - \frac{1}{b}$ to be an integer, then $N = \frac{3b}{a} - \frac{1}{b}$ has to be an integer as well. Since $N = \frac{3b^2 - a}{ab}$, we conclude that $ab$ must divide $3b^2 - a$. This implies that $b$ divides $3b^2 - a$, so $b$ divides $a$.

We can now write $a = k b$. Then $N = \frac{3b - k}{k b}$, so $k b$ divides $3b - k$. $b$ divides $3b - k$, or, finally, $b$ divides $k$. Let $k = l b$. Then $N = \frac{3 - l}{l b}$. We see that $l$ must divide $3 - l$, so $l$ divides $3$. We have $a = l b^2$ where $a$ and $b$ are positive integers, so $l$ is also a positive integer and we have either $l = 1$ or $l = 3$.

If $l = 1$ then $N = \frac{2}{b}$ and $b = 1$ or $b = 2$. In the first case we get $a = 1$, in the second case $a = 4$. If $l = 3$ then $N = 0$ and $a = 3b^2$.

Finally, we need to determine the value of $I$. If $a = b = 1$ then $I = 4$, if $a = 4$ and $b = 2$ then $I = 9$ and if $a = 3b^2$ then $I = 4b^2 = (2b)^2$, so $I$ is indeed a perfect square.

If $a = 4$ and $b = -2$ then $l = 7$. If $a = -4$ and $b = -2$ then $l = 2$. In both cases $l$ is a positive integer, but it is not a perfect square.

Remark. As in the first part of the problem we can show that $a = 4$, $b = -2$ and $a = -4$, $b = -2$ are the only two cases where $l$ is a positive integer but not a perfect square. We know from the first part that at least one of the numbers $a$ and $b$ is negative. As above, we conclude that $a = l b^2$ and $l = b^2 + a + \frac{3 - l}{l b} = (l - 1) b^2 + \frac{3 - l}{l b}$. In this case we can choose $l$ or $b$ to be negative.

We know that $l$ divides $3$. So it can be equal to $1$, $-1$, $3$ or $-3$. If $l = 3$ then $a = 3b^2$ and $l = (2b)^2$ regardless of what $b$ is. When $l = 1$ we have $b = -1$ or $b = -2$. In the first case $l = 0$ and in the second case $a = 4$ and $l = 7$.

If $l = -1$ then $l = -\frac{1}{2}$. So, $l$ is a positive divisor of $4$, but not a perfect square. Hence, $l = 2$, $b = -2$, $a = -4$. In the remaining case we have $l = -3$ and $l = -2b^2 - \frac{1}{2}$. This is an integer when $b = \pm 1$ or $b = \pm 2$, but $l$ is negative in both cases.

Hence, $l$ is a positive integer but not a perfect square if and only if $a = 4$, $b = -2$ or $a = -4$, $b = -2$.