Is there any way of calculating this long formula? $$\left\{\frac{-x^4+y^4-y^2}{\left(x^2+y^2\right)\left(x^2+y^2-1\right)\sqrt{x^2+y^2}\sqrt{x^2+y^2-1}}\right\}\left\{1+\left\{\frac{y^2}{\left(x^2+y^2\right)\left(x^2+y^2-1\right)}\right\}^2\right\}+\left\{\frac{-y^4+x^4-x^2}{\left(x^2+y^2\right)\left(x^2+y^2-1\right)\sqrt{x^2+y^2}\sqrt{x^2+y^2-1}}\right\}^{\:}\left\{1+\left\{\frac{x^2}{\left(x^2+y^2\right)\left(x^2+y^2-1\right)}\right\}^2\right\}-2\frac{y^2}{\left(x^2+y^2\right)\left(x^2+y^2-1\right)}\cdot \frac{x^2}{\left(x^2+y^2\right)\left(x^2+y^2-1\right)}\cdot \frac{-y^4+x^4-x^2}{\left(x^2+y^2\right)\left(x^2+y^2-1\right)\sqrt{y^2+x^2}\sqrt{y^2+x^2-1}}$$
I'm sorry if this question is not proper here, but is there any way(website) to calculate this formula? In principle, this should be zero, but when I do it by hand, it doesn't seem like it. And I want to verify it. Thanks!

Your question is equivalent to asking whether the last term (without the minus sign) and the sum of the first two terms are equal. Expressing these two in the format of one big fraction, the denominators on both sides of the equation are the same: $(x^2+y^2)^{7/2} (x^2+y^2-1)^{7/2}$, therefore comparing the numerators suffices. Make the substitution $p=x^2$ and $q=y^2$ for simplification.
The numerator on the L.H.S. is
$
(-p^2+q^2-q)\big[ (p+q)^2 (p+q-1)^2 + q^2 \big] +(-q^2+p^2-p)\big[ (p+q)^2 (p+q-1)^2 + p^2 \big]
$
which can be simplified into
$-p^3-q^3-(p+q)^3(p+q-1)^2$.
The numerator on the R.H.S. is $2pq(p^2-q^2-p)$.
Notice how all terms when expanded on the R.H.S. must be in the form of $p^m q^n$ where $m,n \in \mathbb{Z} \neq 0$. But on the L.H.S., there are at least a $-p^5$ and $-q^5$ term.
