Can I argue that $g'$ is non zero in this case? Consider two smooth maps $g,f$ given by
$$ {\partial \over \partial x} g(x)= g'(x) = \int_0^1 {\partial \over \partial x} f'(u + t(x-u)) dt = \int_0^1 f''(u + t(x-u)) \cdot t dt  $$ 
where $f' = {\partial f \over \partial x}$ and $u$ is a fixed point in $\mathbb R$.
I want to argue that if $f''$ is non zero then so is $g'$. I tried integration by parts and got
$$ {\partial \over \partial x} g(x) = g'(x) = \int_0^1 f''(u + t(x-u)) \cdot t dt  = {f'(x) \over x - u} - {f(x) - f(u) \over (x-u)^2}$$
but I really don't see why that should be non zero.

Please, does anyone have any ideas on how to argue that $g' \neq 0$?

 A: Suppose $g'(x)=0$ for all $x$. Then 
$$f'(x)=\frac{f(x)-f(u)}{x-u}.$$
One possibility is $f(x)=ax+b$. To see that this is the only possibility, differentiate:
$f''(x)=\frac{f'(x)(x-u)-(f(x)-f(u))}{(x-u)^2}=\frac{(f(x)-f(u))-(f(x)-f(u))}{(x-u)^2}=0.$
Thus $f''(x)=0$ so $f(x)=ax+b$ is the only type of $f$ which makes $g$ identically zero.
A: Okay so OP has left most of the work to others.
From $$g'(x) = \int_0^1 {\partial \over \partial x} f'(u + t(x-u)) dt = \int_0^1 f''(u + t(x-u)) \cdot t dt$$ one can deduce that $f'$ means $\partial f \over \partial x $, $g'$ means $ dg \over dx$ and $f=f(t,x)$,$u$ is a constant.
As OP wrote, use integration by parts and get 
$g'(x) = \int_0^1 f''(u + t(x-u)) \cdot t dt  = {f'(x) \over x - u} - {f(x) - f(u) \over (x-u)^2}$
We want to show that if $f'' \neq 0$ then $g' \neq 0$, which is equivalent to showing that $g'=0 \Rightarrow f''=0$  (basic fact of logic)
Assume $g'(x)=0$, you get $${f(x) \over (x-u)^2} - {f'(x) \over (x-u)}={f(u) \over (x-u)^2}$$ and hence $$ f(x)-f'(x)(x-u)=f(u), \tag{*}$$
Now$(*)$ is a first-order differential equation, you solve it and obtain $f(x)=A(x-u)+f(u)$ where A is an arbitrary constant, thus $f''=0$. By the logic I stated above, you obtain that $g' \neq 0$ provided $f'' \neq 0$
Just an advice: state partial differentiation clearly otherwise it is hard to guess what are you differential with respect for and what are the independent variables!
