1. We are given a function $h(x) = f\left(\frac{f(x) + 2}{3}\right)$ and asked to find $h'(1)$.\n\n2. To find $h'(x)$, we use the chain rule. Let $u = \frac{f(x) + 2}{3}$. Then $h(x) = f(u)$.\n\n3. By the chain rule, $h'(x) = f'(u) \cdot u'$.\n\n4. Next, find $u'$. Since $u = \frac{f(x) + 2}{3}$, we have $$u' = \frac{1}{3} f'(x).$$\n\n5. Therefore, $$h'(x) = f'\left(\frac{f(x) + 2}{3}\right) \cdot \frac{1}{3} f'(x).$$\n\n6. We need to evaluate $h'(1)$, so substitute $x=1$: $$h'(1) = f'\left(\frac{f(1) + 2}{3}\right) \cdot \frac{1}{3} f'(1).$$\n\n7. From the graph, observe the values: $f(1) = 0$ (since the black curve crosses the x-axis at $x=1$).\n\n8. Also from the graph, $f'(1)$ is the slope of $f$ at $x=1$. The black curve has a local minimum near $x=1$, so $f'(1) = 0$.\n\n9. Calculate the inner argument: $$\frac{f(1) + 2}{3} = \frac{0 + 2}{3} = \frac{2}{3}.$$\n\n10. Find $f'\left(\frac{2}{3}\right)$. From the graph, at $x=\frac{2}{3}$, the slope $f'(x)$ is positive but we do not need the exact value because $f'(1) = 0$ will multiply it.\n\n11. Since $h'(1) = f'\left(\frac{2}{3}\right) \cdot \frac{1}{3} \cdot 0 = 0$, the derivative $h'(1) = 0$.\n\n**Final answer:** $h'(1) = 0$.