Is it true that $\frac{f(b)-f(a)}{b-a}-\frac{b-a}{g(b)-g(a)}=f'(c)-\frac{1}{g'(c)}$

Consider a composite function $f\circ g(x)$ Hence we have $$ (f\circ g)'(x)= f '(g(x)) g'(x) $$

That is, when $f\circ g(x)=x$, then $\frac{1}{g'(x)} =f'(g(x))$

In general we guess that $f'(g(x)) \neq f'(x)$

Consider $f(x)=x^2,\ g(x)=\sqrt{x}$ on $[0,1]$, which is a counterexample : $\frac{f(0)-f(1)}{0-1} = \frac{g(0)-g(1)}{0-1}=1$

Further, $f'(x)=2x,\ g'(x)= \frac{1}{2\sqrt{x}}$ so that $2x-2\sqrt{x}\neq 0$ for all $x\in (0,1)$

Consider two functions $f$, $g$ such that $f'(x)=1/g'(x)$. Then the RHS is always zero, but not necessarily the LHS.

Example: Let $g(x):=x^3$, $f'(x)=\frac{1}{3x^2}$, so $f(x)=-\frac{1}{3x}$. Take $a=1$, $b=2$, to get $$\frac{f(b)-f(a)}{b-a}-\frac{b-a}{g(b)-g(a)}=\frac{1}{42}\ne0=f'(c)-\frac{1}{g'(c)}$$ for any $c\in[1,2]$.