I was searching for the answer to this question here and found a very specific example listed: $$f(x) = \frac{x-3}{x^3}$$
None of the answers I found to the question answered how to determine if this function is one-to-one or not. Checking with the horizontal line test on a graphing calculator, the function is indeed not one-to-one, but I want to know how to do this algebraically.
You only need to find a pair of nonzero numbers $x$ and $y$ such that $x\ne y$ but $f(x)=f(y)$. Given that $x\ne y$ and $x,y\ne0$, \begin{align} f(x)=f(y) &\Leftrightarrow xy^3-3y^3=yx^3-3x^3\\ &\Leftrightarrow xy(y^2-x^2)=3(y^3-x^3)\\ &\Leftrightarrow xy(y+x)=3(y^2+yx+x^2)\\ &\Leftrightarrow (y-3)x^2+y(y-3)x-3y^2=0.\tag{1}\\ \end{align} This is a quadratic equation in $x$. Its discriminant $\Delta=y^2(y-3)^2+12y^2(y-3)=y^2(y-3)(y+9)$. If $\Delta>0$, then the quadratic equation has two distinct roots. Hence one of them must be different from $y$. E.g. if we pick $y=4$, then $\Delta=52>0$ and equation $(1)$ has two distinct roots $x=-2(1\pm\sqrt{13})$. Hence $f\big(-2(1+\sqrt{13})\big)=f\big(-2(1-\sqrt{13})\big)=f(4)$ and we conclude that $f$ is not one-to-one.