A problem about the differential mean value theorem $2ηf(1)+(c^2-1)f'(η)=f(ξ)$

Let $g(x)= x^2 f(1)+(c^2-1) f(x)$.

Then $g(1)-g(0)=c^2 f(1)+(1-c^2) f(0)$ so by Rolle there exists $\eta \in (0,1)$ such that: $$g'(\eta)=2 \eta f(1)+(c^2-1) f'(\eta)=\frac{g(1)-g(0)}{1-0}=c^2 f(1)+(1-c^2) f(0)$$ Moreover, by the mean value theorem, as $f$ is continuous and $c^2 f(1)+(1-c^2) f(0) \in [f(0),f(1)]$, there exists $\xi \in [0,1]$ such that $$f(\xi)=c^2 f(1)+(1-c^2) f(0)$$ And so: $$2 \eta f(1)+(c^2-1) f'(\eta)=c^2 f(1)+(1-c^2) f(0)=f(\xi)$$