The curve $y=a{x}^3+b{x}^2+cx$ is inclined at $45°\text{ to }x\text{-axis }$ at (0,0) but it touches x-axis at  (1,0), then 

$\frac{dy}{dx}=3a{x}^2+2bx+c=\tan 45°=1\text{ at }x=0$

$\therefore c=1$

Now ,${\left[\frac{dy}{dx}\right]}_{(1,0)}=3a+2b+c=0$ as $x$ -axis is tangent. 

$\therefore 3a+2b+1=0 [\because c=1]\cdots \cdots \cdots \cdots \cdots \left(i\right)$

Now, (1, 0) lies on curve, 

$a+b+1=0$

Solving (i) and (ii), we get $a$= 1, b = -2 

$\therefore f\left(x\right)=x^3-2x^2+x$

and $f\text{'}\left(x\right)=3x^2-4x+1\Rightarrow f\text{'}\text{'}\left(x\right)=6x-4$