Let $a,b \in \mathbb{N}$ and $a+\frac{b}{a}-\frac{1}{b}=k \in \mathbb{Z}$. From $a^2b+b^2-a=kab$, we get that $b|a$. Let $a=bq$, $q>0$, $q \in \mathbb{Z}$. Then $b^3q^2+b^2-bq=kb^2q$, and after dividing with $b$ ($b>0$) we have $b^2q^2+b-q=kbq$, therefore $q|b$. Let $b=qt$, $t>0$, $t \in \mathbb{Z}$. Then $q^4t^2+qt-q=kq^2t$, and after dividing with $q$ ($q>0$), we have $q^3t^2+t-1=kqt$, therefore $t|1$ and $t>0$, so $t=1$. Since $b=qt$ and $t=1$, we have $b=q$. From $a=bq$ and $b=q$, we have $a=b^2$. Finally $k=a+\frac{b}{a}-\frac{1}{b}$ and $a=b^2$ we have $k=b^2+\frac{b}{b^2}-\frac{1}{b}=b^2$.