If $f$ and $g$ are differentiable functions in $(0,1)$ satisfying $f(0)=2=g(1)$, $g(0)=0$ and $f(1)=6$, then for some $c\in (0,1)$:

Let's define a new function $h(x)=f(x)-2g(x)$.Given that $f(x)$ and $g(x)$ are differentiable (and thus continuous), $h(x)$ is also continuous on $[0,1]$ and differentiable on $(0,1)$.Calculate the values at the boundaries:
$$h(0)=f(0)-2g(0)=2-2(0)=2$$
$$h(1)=f(1)-2g(1)=6-2(2)=6-4=2$$
Since $h(0)=h(1)$, $h(x)$ satisfies the conditions of Rolle’s Theorem.According to Rolle’s Theorem, there exists at least one $c\in (0,1)$ such that $h^{\prime }(c)=0$.
$$h^{\prime }(c)=f^{\prime }(c)-2g^{\prime }(c)=0$$
$$⟹f^{\prime }(c)=2g^{\prime }(c)$$