$\text{ If }f(x+y)-f\left(x\right)-f\left(y\right)=\left|x\right|y+x{y}^2\mathrm{\forall }x,y\in R\text{ and }{f}^{\mathrm{\prime }}\left(0\right)=0\text{ then }$

$\begin{array}{l}{f}^1\left(x\right)=L_{x\to 0}\frac{f(x+h)-f\left(x\right)}{h}=L_{h\to 0}\frac{f\left(h\right)+\left|x\right|h+x{h}^2}{h}\\ ={\mathrm{Lt}}_{h\to 0}\left(f^1\left(h\right)+\left|x\right|+xh\right)=\left|x\right|\end{array}$