This question already has an answer here:
Note that $f'(x) = h(x)$, $g'(x)=f(x)$, and $h'(x)=g(x)$. Then consider
$$\frac{d}{dx} [f^3 + g^3 + h^3 – 3 f g h]$$
and show that it is zero using the above derivatives. Viz.,
$$\begin{align}\frac{d}{dx} [f^3 + g^3 + h^3 – 3 f g h] &= 3 f^2 f’+3 g^2 g’+3 h^2 h’ – 3 f’gh-3 f g’h – 3 f g h’\\ &= 3 f^2 h+3 g^2 f + 3 h^2 g – 3 f’gh-3 f g’h – 3 f g h’\\ &= 0 \\\end{align}$$
Also note that $f(0)=1$, $g(0)=0$, and $h(0)=0$. Thus the integration constant of the above equation is 1, and
$$f^3(x) + g^3(x) + h^3(x) – 3 f(x) g(x) h(x)=1$$
Notice that $h’=g$ and $g’=f$ and $f’=h$, now consider $$(fgh)’=f(g’h+gh’)+f’gh$$ $$=fg’h+fgh’+f’gh$$ $$=f^2h+fg^2+gh^2$$ whcih implies that $$fgh=\int f^2h+fg^2+gh^2$$$$=\frac{f^3}{3}+\frac{g^3}{3}+\frac{h^3}{3}+c$$ which implies that $$f^3(x)+g^3(x)+h^3(x)-3f(x)g(x)h(x)=c$$ for all $x\in \mathbb R$, substitute x=0 to get that c=1 and hence you get the result.