jmc

Let $y=10^x.$ Then $$10^x-100^x=y-y^2=\frac{1}{4}-{\left(y-\frac{1}{2}\right)}^2.$$Thus, the maximum value is $\boxed{\frac{1}{4}},$ which occurs when $y=\frac{1}{2},$ or $x={\log }_{10}\left(\frac{1}{2}\right).$