A follow up to a previous question: Chain rule when sum is differentiable but individual functions are not
Take some $g_1, g_2, h_1,h_2: \mathbb{R}\to \mathbb{R}$ that are everywhere twice differentiable and and always have a strictly positive derivative. Take also some $f_1,f_2: \mathbb{R}\to \mathbb{R}$ which need not be differentiable. However, we know that for some $x_0$: $$ g_1'(f_1(x_0)) = h_1'(f_1(x_0)) >0 $$ $$ g_2'(f_2(x_0)) = h_2'(f_2(x_0)) >0 $$ $$ \frac{d}{dx}[g_1(f_1(x))+ g_2(f_2(x))] \big|_{x=x_0} = 0 $$ (We only know that the above derivative exists at $x_0$, not everywhere). Can we say the following? $$ \frac{d}{dx}[h_1(f_1(x))+ h_2(f_2(x))] \big|_{x=x_0} = 0 $$ (We do not assume that the derivative right above exists).
Let $f_1(x)=x^{\frac{1}{3}}$, $f_2(x)=-x^{\frac{1}{3}}$, let $g_1(x)=g_2(x)=x$, and let $h_1(x)=x+x^3$, $h_2(x)=x$.
Then certainly $g_1$, $g_2$, $h_1$, $h_2$ are twice differentiable with strictly positive derivatives which coincide at $f_1(0)=f_2(0)=0$.
Moreover, we have $$g_1(f_1(x))+g_2(f_2(x))= x^{\frac{1}{3}}-x^{\frac{1}{3}}=0$$ everywhere, so certainly the conditions of the question are satisfied.
However, $$h_1(f_1(x))+h_2(f_2(x)) = x^{\frac{1}{3}}+ x + (-x^\frac{1}{3})=x,$$ so $$\frac{d}{dx}[h_1(f_1(x))+ h_2(f_2(x))] \big|_{x=0} = 1.$$