Show that $f$ is bijective.

If you want to show the injectivity, here is one way: $$\frac{2x-s}{x(s-x)} = \frac{2y-s}{y(s-y)}$$ Bringing to a common denominator, we see that $$(2x-s)y(s-y) - (2y-s)x(s-x) = 0$$ Assume $x\ne y$ but $x>y$, then $$(2x - s)y(s-y) - (2y-s)x(s-\color{red}y)<0 \\ \left[(2x-s)y - (2y-s)x\right](s-y) < 0$$ Drop the factor $(s-y)$ since it is positive: $$(2x-s)y - (2y-s)x < 0 \\ -sy + sx < 0 \\ x < y$$ Contradiction. Since the case $x < y$ is similar, we must have $f(x)=f(y)\implies x=y. \ \ \square$

ANOTHER APPROACH:

Observe that $$f'(x)=\dfrac{2x(s-x)-(2x-s)(s-2x)}{x^2(s-x)^2}=\frac{2x^2-2xs+s^2}{x^2(s-x)^2}=\frac{(x-s)^2+x^2}{x^2(s-x)^2}>0.$$ So, since $f$ is derivable on the interval $(0,s)$, there the derivative is positive, the function is stricktly increasing

Moreover, $\lim_{x\to0^+} f(x)=-\infty$ and $\lim_{x\to s^-}f(x)=+\infty$.

So $f$ maps bijectively $(0,s)$ into $\Bbb R$.