I'm having problems showing this result:
Let $f:U\rightarrow\mathbb{R}^m$ be a Lipschitz function with $a\in U$ and $g:V\rightarrow\mathbb{R}^p$, $V$ open, with $f(U)\subset V$ and $b=f(a)$. If $g'(b)=0$ then $g\circ f:U\rightarrow \mathbb{R}^p$ is differentiable in $a$, with $(g\circ f)'(a)=0$.
I have tried to use the definition of $g$ being differentiable in $b=f(a)$ and using $f(v)$ as the increment $h$ so:
$g(f(a)-f(v))-g(f(a))=g'(f(a))\cdot f(v)+r(f(a),f(v))$
Since $g'(b)=g'(f(a))=0$, we have:
$g(f(a)-f(v))-g(f(a))=r(f(a),f(v))$
I have tried to do through the inverse way but couldn't do it either. Can someone help me?