Proving that function with domain (-1,1) is injective. Function $g\colon (-1,1) \rightarrow \mathbb R$ is defined by $g(x)=\dfrac{x}{1-x^2}$. Prove that $g(x)$ is injective.
Work: 
I shifted the equation so that it ends up like $\dfrac{x}{y}=\dfrac{1-x^2}{1-y^2}$. From here, I thought about using cases such as $x>y$ or $y>x$ to prove it, but it does not seem to work. I tried to manipulate the equation into various forms, but I cannot end up with $x=y$. 

Rewrite your function as follows $f(x) = \frac{x}{1-x^2} = \frac{1}{\frac{1}{x} - x}$
Suppose $f(x) = f(y)$. That is
$$ \frac{1}{\frac{1}{x} - x} = \frac{1}{\frac{1}{y} - y}  \iff  \frac{\frac{1}{y}-y}{\frac{1}{x} - x} = 1$$
We want to show $x = y$. Suppose not. Then we either have $x > y$ or $y > x$.
Suppose $y > x$. Then $-y < -x $ and $\frac{1}{y} < \frac{1}{x} $. Hence
$$ 1 =  \frac{\frac{1}{y}-y}{\frac{1}{x} - x} <  \frac{\frac{1}{x}-x}{\frac{1}{x} - x} = 1$$
Contradiction!
Similarly, I let you do it the case $x  > y$ which will lead you to a contradiction.
Therefore $x = y$ which implies $f$ is injective
