jmc

The graph of $$x^2-xy={\left(x-\frac{y}{2}\right)}^2-\frac{y^2}{4}$$is a parabola with vertex at $\left(\frac{y}{2},-\frac{y^2}{4}\right).$

We divide into cases, based on the value of $y.$

If $y\le 0,$ then $$|x^2-xy|=x^2-xy$$for $0\le x\le 1.$ Since $x^2-xy$ is increasing on this interval, the maximum value occurs at $x=1,$ which is $1-y.$

If $0\le y\le 1,$ then $$|x^2-xy|=\{\begin{array}{cl}xy-x^2& \text{for 0≤x≤y},\\ x^2-xy& \text{for y≤x≤1}.\end{array}$$Thus, for $0\le x\le y,$ the maximum is $\frac{y^2}{4},$ and for $y\le x\le 1,$ the maximum is $1-y.$

If $y\ge 1,$ then $$|x^2-xy|=xy-x^2$$for $0\le x\le 1.$ If $1\le y\le 2,$ then the maximum value is $\frac{y^2}{4},$ and if $y\ge 2,$ then the maximum value is $y-1.$

For $y\le 0,$ the maximum value is $1-y,$ which is at least 1. For $1\le y\le 2,$ the maximum value is $\frac{y^2}{4},$ which is at least $\frac{1}{4}.$ For $y\ge 2,$ the maximum value is $y-1,$ which is at least 1.

For $0\le y\le 1,$ we want to compare $\frac{y^2}{4}$ and $1-y.$ The inequality $$\frac{y^2}{4}\ge 1-y$$reduces to $y^2+4y-4\ge 0.$ The solutions to $y^2+4y-4=0$ are $-2\pm 2\sqrt{2}.$ Hence if $0\le y\le -2+2\sqrt{2},$ then the maximum is $1-y,$ and if $-2+2\sqrt{2}\le y\le 1,$ then the maximum is $\frac{y^2}{4}.$ Note that $1-y$ is decreasing for $0\le y\le -2+2\sqrt{2},$ and $\frac{y^2}{4}$ is increasing for $-2+2\sqrt{2}\le y\le 1,$ so the minimum value of the maximum value occurs at $y=-2+2\sqrt{2},$ which is $$1-(-2+2\sqrt{2})=3-2\sqrt{2}.$$Since this is less than $\frac{1}{4},$ the overall minimum value is $\boxed{3-2\sqrt{2}}.$