Let $f: \mathbb{R} \rightarrow \mathbb{R} $ be continuously differentiable, then it's well know that
\begin{align*} f(x) & = f(y) + \int_0^1 \frac{\partial f (y- t(x-y))}{\partial t} \, dt \\ & = f(y) + \int_0^1 f' (y- t(x-y)) (x-y) \, dt \end{align*}
Let $g: \mathbb{R} \rightarrow \mathbb{R} $ also be continuously differentiable. I perform the following derivation:
\begin{align*} g(f(x)) & = g(f(y)) + \int_0^1 \frac{\partial g( f (y- t(x-y)))}{\partial t} \, dt \\ & = g(f(y)) + \int_0^1 g'(f (y- t(x-y) ) f' (y- t(x-y) (x-y) \, dt \tag{1} \end{align*}
I thought this derivation was correct but taking $g(f(x)) := f(x) / x^2$ and $f(x) := x^3$ I calculate $$g'(f(x)) = \frac{f'(x) x^2 - 2x f(x)} { x^4} = \frac{3x^4 - 2x^4}{x^4} =1$$ and $f'(x) = 3x^2$ and plugging this into my derivation gives
\begin{align*} x & = y + \int_0^1 (y- t(x-y))^2 (x-y) \, dt \end{align*} and this is just wrong. Where is the error in my derivation of the formula in $(1)$ and how could I fix it?
There is a mess about our $g$ function. I guess you want to say $ g(x)=x $, but if $ g(x)=x $, then $ g(f(x))= f(x) $ rather than $f(x)/{x^2}$. You need a clearer definition for $g$.