Prove: If $x+y+z=xyz$ then $\frac {x}{1-x^2} +\frac {y}{1-y^2} + \frac {z}{1-z^2}=\frac {4xyz}{(1-x^2)(1-y^2)(1-z^2)}$ If $x+y+z=xyz$, prove that:
$$\frac {x}{1-x^2} +\frac {y}{1-y^2} + \frac {z}{1-z^2}=\frac {4xyz}{(1-x^2)(1-y^2)(1-z^2)}$$.
My Attempt:
$$L.H.S=\frac {x}{1-x^2}+\frac {y}{1-y^2}+\frac {z}{1-z^2}$$
$$=\frac {x(1-y^2)(1-z^2)+y(1-x^2)(1-z^2)+z(1-x^2)(1-y^2)}{(1-x^2)(1-y^2)(1-z^2)}$$
$$=\frac {x+y+z-xz^2-xy^2+xy^2z^2-yz^2-yx^2+x^2yz^2-zy^2-zx^2+zx^2y^2}{(1-x^2)(1-y^2)(1-z^2)}$$.
I could not move on from here.Please help.
Thanks  

Continuing from where you left, expressing terms of the numerator as:
$$-xz^2-x^2z+x^2yz^2 =-xz (z+x-xyz) =-xz (-y) $$
$$-xy^2-x^2y+x^2y^2z =-xy (x+y-xyz) =-xy (-z) $$
$$-yz^2-y^2z+xy^2z^2 =-yz (y+z-xyz)=-yz (-x) $$
$$x+y+z=xyz $$
Now add everything and the result follows. Hope it helps. 
