If $x+y+z=xyz$, prove $\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$ 
If $x+y+z=xyz$, prove
$\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$
given that $x^2~,~y^2~,~z^2\ne1$

I came across this question in an ancient ($19$th century) Trigonometry book, and this is the method they use to prove the result (please note: I understand this method fully):
Let $x=\tan A$ ,$~y=\tan B$ and $z=\tan C$ which is acceptable without any loss of generality.
This means that we are saying that
$$\tan A+\tan B+\tan C=\tan A\tan B\tan C$$
Consider $\tan (A+B+C)$:
$$\tan (A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1- \tan A \tan B- \tan C \tan A - \tan B \tan C}$$
So if $~\tan A+\tan B+\tan C=\tan A\tan B\tan C~~$ then $~~\tan (A+B+C)=0$. Hence, let $~A+B+C=\pi$.
Now consider $\tan (2A+2B+2C)$:
$$\tan (2A+2B+2C)=\frac{\tan 2A+\tan 2B+\tan 2C-\tan 2A\tan 2B\tan 2C}{1- \tan 2A \tan 2B- \tan 2C \tan 2A - \tan 2B \tan 2C}=0$$
$$\implies \tan 2A+\tan 2B+\tan 2C=\tan 2A\tan 2B\tan 2C$$
$$\implies \frac{2\tan A}{1-\tan^2 A}+\frac{2\tan B}{1-\tan^2 B}+\frac{2\tan C}{1-\tan^2 C}=\frac{2\tan A}{1-\tan^2 A}\times\frac{2\tan B}{1-\tan^2 B}\times\frac{2\tan C}{1-\tan^2 C}$$
$$\therefore\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$$
as required.

My question is, is there any other way of proving this without this rather heavy use of trigonometry? I also would prefer not to work through heaps of algabraic manipulation and expansion to obtain the required result, although if that's necessary I will put up with it ;)
Thank you for your help.

Let $z=0$.
Thus, $x+y=0$ and
$$\sum_{cyc}\frac{2x}{1-x^2}=\frac{2x}{1-x^2}-\frac{2x}{1-x^2}=0=\prod_{cyc}\frac{2x}{1-x^2}.$$
Now, let $xyz\neq0.$
Thus, we need to prove that:
$$\sum_{cyc}\frac{1}{\frac{2x}{1-x^2}\cdot\frac{2y}{1-y^2}}=1.$$
Indeed, $$\sum_{cyc}\frac{1}{\frac{2x}{1-x^2}\cdot\frac{2y}{1-y^2}}=\sum_{cyc}\frac{(1-x^2)(1-y^2)}{4xy}=\sum_{cyc}\frac{(\frac{xyz}{x+y+z}-x^2)(\frac{xyz}{x+y+z}-y^2)}{4xy}=$$
$$=\sum_{cyc}\frac{(yz-x(x+y+z))(xz-y(x+y+z))}{4(x+y+z)^2}=$$
$$=\sum_{cyc}\frac{x^2yz-(x+y+z)(x^2y+x^2z)+xy(x+y+z)^2}{4(x+y+z)^2}=$$
$$=\sum_{cyc}\frac{\frac{1}{3}xyz-x^2y-x^2z+xy(x+y+z)}{4(x+y+z)}=\sum_{cyc}\frac{\frac{1}{3}xyz+xyz}{4xyz}=1.$$
