The question is this: If $f$, $g$, $h$ are continuous functions on $[a,b]$ which are differentiable on $(a,b)$ then prove that there exists $c \in (a,b)$ such that $f(a)[g(b)h^{\prime}(c) - h(b)g^{\prime}(c)]+h(a)[f(b)g^{\prime}(c) - g(b)f^{\prime}(c)] = g(a)[f(b)h^{\prime}(c) - h(b)f^{\prime}(c)]$.
I tried to work it backwards by deducing something from this result and see what $c$ needs to satisfy. As you can check, the result is equivalent to
\begin{equation}(f(a)-f(b))[g(b)h^{\prime}(c) - h(b)g^{\prime}(c)] + (h(a)-h(b))[f(b)g^{\prime}(c) - g(b)f^{\prime}(c)] = (g(a) - g(b))[f(b)h^{\prime}(c) - h(b)f^{\prime}(c)]\end{equation}
By the mean value theorem, there exists $x_1,x_2,x_3 \in (a,b)$ such that $f(a)-f(b) = (a-b)f^{\prime}(x_1)$, $g(a)-g(b) = (a-b)g^{\prime}(x_2)$ and $h(a)-h(b) = (a-b)h^{\prime}(x_3)$. After substituting them back and some cancelling, we have
\begin{equation}f^{\prime}(x_1)[g(b)h^{\prime}(c) - h(b)g^{\prime}(c)] + h^{\prime}(x_3)[f(b)g^{\prime}(c) - g(b)f^{\prime}(c)] = g^{\prime}(x_2)[f(b)h^{\prime}(c) - h(b)f^{\prime}(c)]\end{equation}
Then I realize this is not something easy to work with, and I am seeking an alternative strategy to approach this problem. Maybe directly dealing with the question of what $c$ might be and getting properties from it would be a better way?