$f,g : [0,∞)→[0,∞)$ are respectively non-increasing and non-decreasing functions. Let $h(x):=g(f(x))$. It is given that $f,g$ are derivable in their domain and that $h(0)=0$.
Now $f'(x)≤0$ and $g'(x)≥0$ for all $x$ in their domain. $h'(x)=g'(f(x))f'(x)≤0$ for all $x$. So $h$ is non-increasing. Since the range of $h$ is $[0,∞)$ and $h(0)=0$, it follows that $h$ is identically zero. Is this correct?
Looks right to me.
To be honest, I don't know why you even need the condition about differentiability.
If $x\le y$ implies $f(x)\ge f(y)$, i.e. $f(y)\le f(x)$, which in turn implies $g(f(y))\le g(f(x))$, i.e. $h(x)\ge h(y)$, you can see straight away that $h$ is non-increasing.
Here is a calculation that shows a bit better what $f$ and $g$ look like. Let $f(0)=c\ge 0$. Because $0=h(0)=g(f(0))=g(c)$ and $g$ is non-decreasing, you also have $g(x)=0$ for all $x\in[0,c]$. But now, $f$ is non-increasing, so for all $x\in[0,\infty)$ you have $f(x)\le f(0)=c$, i.e. $f(x)\in[0,c]$ and so $h(x)=g(f(x))=0$.