Which of the following functions have the graph symmetrical about the origin?

$f\left(x\right)+f\left(y\right)=f\left(\frac{x+y}{1-xy}\right)$

Replacing y by –x, we get f(x) + f(–x) = f(0).....(1)
Putting x = y = 0, we get f(0) + f(0) = f(0) or f(0) = 0

$\therefore$ f(x)+f(–x)=0 Hence, f(x) is an odd function.

$f\left(x\right)+f\left(y\right)=f\left(x\sqrt{1-y^2}+y\sqrt{1-x^2}\right)$

Replacing y by –x, we get f(x)+f(–x)=f(0)
Putting x=y=0 we get f(0)+f(0)=f(0) or f(0)=0

$\therefore$f(x)+f(–x)=0. Hence, f(x) is an odd function.
f(x+y)=f(x)+f(y) Replacing y by –x,
we get f(0)=f(x)+f(–x) Putting x=y=0,
we get f(0+0)=f(0)+f(0) or f(0)=0

$\therefore$f(x)+f(–x)=0. Hence, f(x) is an odd function.