The value of x for which $\left(x-x^2\right)$ is maximum, is:

Let $f\left(x\right)=x-x^2$ 
Therefore, $f^{\mathrm{\prime }}\left(x\right)=1-2x$
Taking $f^{\mathrm{\prime }}\left(x\right)=0,\text{ we get, }x=\frac{1}{2}$
Now, $f^{\prime \prime }\left(x\right)=-2$
At $x=\frac{1}{2},f^{\prime \prime }\left(x\right)<0$
Therefore, $f\left(x\right)$ is maximum at $x=\frac{1}{2}$.