Wow what a crazy thing

Easy one,

First of all we'll try generalization.

So $y = x^n \times {1-x}^m$

So $\dfrac{dy}{dx} = nx^{n-1} \times (1-x)^m - m(1-x)^{m-1} x^n$

Now we know for Maxima or Minima , We have $\dfrac{dy}{dx} = 0$

So We get $0 = nx^{n-1} \times (1-x)^m - m(1-x)^{m-1} x^n$

So On simplyfying we get,

$\dfrac{m}{n} (1-x) = x$

or , $n - nx = mx$

OR $x = \dfrac{n}{m+n}$

So General Term for $\boxed{x= \dfrac{n}{m+n}}$

SO for this problem we get $m = \dfrac{1}{2}$ and $n = 1$

Hence Putting m and n we get,

$x = \dfrac{2}{3} = 0.666...$

So Value of x such that $y = x^1 \times {1-x}^{\dfrac{1}{2}}$ in minimum will be

$\boxed{x = \dfrac{2}{3} = 0.666...}$