= Solution
Let $L(f)=\mathcal T[f]-f'(1/3)$. Direct substitution shows $L(1)=L(x)=L(x^2)=0$, so the functional annihilates all quadratic polynomials. Taylor's integral-remainder formula is
$$
f(x)=f(0)+xf'(0)+\frac{x^2}{2}f''(0)
+\int_0^1\frac{(x-t)_+^2}{2}f'''(t)\,dt.
$$
Applying $L$ and interchanging the finite evaluations and derivative with the continuous integral gives the <Peano kernel> representation $L(f)=\int_0^1K(t)f'''(t)dt$, where
$$
K(t)=\frac23(1/2-t)_+^2+\frac16(1-t)^2-(1/3-t)_+
=\begin{cases}
5t^2/6,&0\le t\le1/3,\\
5t^2/6-t+1/3,&1/3\le t\le1/2,\\
(1-t)^2/6,&1/2\le t\le1.
\end{cases}
$$
All three expressions are nonnegative; the middle quadratic has positive leading coefficient and discriminant $-1/9$. Therefore the best possible norm bound has constant $\int_0^1K(t)dt$. Evaluate that integral particularly simply by using $f(x)=x^3/6$, for which $f'''=1$:
$$
\int_0^1K(t)dt=L(x^3/6)=\frac1{12}-\frac1{18}=\frac1{36}.
$$
It follows that
$$
\boxed{|\mathcal T[f]-f'(1/3)|\le\frac1{36}\|f'''\|_\infty,\qquad c_{\min}=\frac1{36}.}
$$
The same cubic attains equality, so the constant is sharp, not merely sufficient. This is the <sharp Peano bound for an interior three-point first derivative>.
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