Prove that $\exists x_0, x_1\in (0,1)$, such that $\frac{f'(x_0)}{x_0}+\frac{f'(x_1)}{x_1^2}=5$

Let $f:[0,1]\to\mathbb{R}$ be a differentiable function, such that $f(0)=0$ and $f(1)=1$. Prove that there exist different $x_0, x_1\in (0,1)$, such that $$\frac{f'(x_0)}{x_0}+\frac{f'(x_1)}{x_1^2}=5$$

I have thought of possibly using Cauchy's theorem, so that we would only need to prove that there exist $k, l\in (0,1)$ such that $f(k)=k^2$ and $f(l)=l^3$, but I don't know how to prove these. Any hint?

Edit 1: Apparently my thought here is wrong, so any ideas?

Consider the function $g(x) = \frac{1}{3}x^3$. Then as it's differentiable on $(0,1)$ and continuous on $[0,1]$, by Cauchy Mean Value Theorem we have $\exists c \in (0,1)$, s.t.:

$$\frac{f(1) - f(0)}{g(1) - g(0)} = \frac{f'(c)}{g'(c)} \implies \frac{f'(c)}{c^2} = 3$$

Similarly now taking $h(x) = \frac{1}{2}x^2$, by Cauchy Mean Value Theorem $\exists d \in (0,1)$, s.t.:

$$\frac{f(1) - f(0)}{h(1) - h(0)} = \frac{f'(d)}{h'(d)} \implies \frac{f'(d)}{d} = 2$$

Hence the proof.