Given $g(x)=-x\sin^2(\frac{1}{x})$ for $0<x\leq1$
My attempt: let fixed point given by $g(x)-x=-x\sin^2(\frac{1}{x})-x=0$
$$0=-x\left(\sin^2\left(\frac{1}{x}\right)+1\right) $$ Therefore only for $x=0$ will make the entire term $0$. Because the latter term is bounded below by one and is monotonically increasing from $\sin^2(0)$ to $\sin^2(1)$.
The function $g(x)=-x\sin^2(\frac{1}{x})$ has NO fixed point in the interval $(0,1]$. The continuous extension of $g$ to $[0,1]$ has just one fixed point in $[0,1]$, i.e. $x=0=\lim_{x\to 0}g(x)=0$ (here we need the fact that $\sin^2(\frac{1}{x})$ is bounded).
In order to show the first part we don't need to the monotone property (which is false in this case) or the bounded property. The equation $$0=-x\left(h^2(x)+1\right)$$ has NO solutions in $(0,1]$ because because $x>0$ and $h^2(x)\geq 0$ implies $h^2(x)+1\geq 1>0$ for ANY function $h$ defined in $(0,1]$