Maximum value of Exponent function

Rewrite the function as $f(x)=\frac{x}{2^x}-3\left(\frac{x}{2^x}\right)^2$

Letting $u=\frac{x}{2^x}$ , we can define $f(x)=g(u)=u-3u^2=\frac{1}{12}-3\left(u-\frac16\right)^2$

The maximum value of $g(u)$ is clearly $\frac{1}{12}$ , when $u=\frac16$ .

So we just need to check that the function $h(x)=\frac{x}{2^x}-\frac16$ has a real root. $h(x)$ is certainly continuous, and defined for all $x$ ; also $h(0)<0$ and $h(1)>0$ so there is indeed a root and the maximum value of $f(x)$ is also $\boxed{\frac{1}{12}}$ .