What's the fastest way to prove if a function has an inverse?

Question

I have been searching online for a method that shows whether or not a function has an inverse and the fastest method I can find is that you can prove that $f(x)=f(y)\Longrightarrow x=y$ which proves that the function is one-to-one. For the function $y=\frac{x}{x^2+1}$ specifically, what is the fastest way to prove that this function doesn't have an inverse.

Answer 1

Your function $f(x)=\frac{x}{x^2+1}$ is of course continuous, since it is a ratio of polynomials and the denominator has no real roots. If a continuous function is a bijection (i.e. it has an inverse), then it must be monotonic (see this question for a proof). But $f(x)$ is clearly not monotonic, since $f(0)<f(1)$ but $f(2)<f(1)$.

Answer 2

$f(\frac 1x)=\dfrac{\frac 1x}{\frac 1{x^2}+1}=\dfrac{x^2}{x(1+x^2)}=\dfrac{x}{1+x^2}=f(x)\quad$ thus $f$ is not injective on whole $\mathbb R$.

You need to restrict the function to intervals $(-\infty,-1),[-1,1],(1,+\infty)$ to recover injectivity (that need to be confirmed by studying $f'(x)$ sign on these intervals and show $f$ is strictly monotonic on these).