= Solution
Differentiate the integral formula using the <Leibniz integral rule>. Its two moving-endpoint terms cancel, leaving
$$
y'(x)=\frac2{3x^3}\int_0^x\zeta f(\zeta)d\zeta-\frac13\int_x^\infty\frac{f(\zeta)}{\zeta^2}d\zeta.
$$
Each term before subtraction lies between zero and $m/(3x)$. Therefore
$$
\boxed{-\frac m{3x}\leq y'(x)\leq\frac m{3x}.}
$$
The bounds are sharp at any fixed positive $x$: put $f=m$ below that point and zero above it for the upper bound, or reverse those choices for the lower bound. If continuity of the forcing is required, smooth approximations approach the same bounds.
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