We note that $$ |\frac{x^2f(x)-y^2f(y)}{x-y}| \leq |x-y|$$which by definition of continuity and differentiability gives that $g(x)=x^2f(x)$ is continuous and differentiable on $\mathbb{R}^+$. As $1/x^2$ is differentiable on $\mathbb{R}^+$ we can say that $g(x)/x^2=f(x)$ is differentiable. Thus $$(x^2f(x))'=0 \implies f(x)=\frac{c}{x^2}$$for some arbitrary constant $c>0$

This is basically the old Russian coffin problem.