$f: [0, 1] \to \mathbb{R}$ is continuous. Prove existence of $c \in (0, 1)$ where $\int_{0}^{1} f(x) x^2 dx = \frac{f(c)}{3}$

Question

My approach with partial integration:

$$\int_{0}^{1} f(x) x^2 dx = \frac{f(1)}{3} - \int_{0}^{1} f'(x) x^2 dx = \frac{f(c)}{3}$$

By mean value theorems, there exists $d \in [0, 1]$ such that:

$$\int_{0}^{1} f'(x) x^2 dx = f'(d) \cdot d^2$$

Hence:

$$\frac{f(1)}{3} - f'(d) \cdot d^2 = \frac{f(c)}{3}$$

This is where I can not go further. If $d$ is a maximal or a minimal of $f$ in $(0, 1)$, then $f(1) = f(c)$ but $c$ can not be equal $0$ or $1$.