Write $y = kx$, with $k$ a positive real number. We want to find the minimum of the functions $f(x) = kx²$, $g(x) = kx^2-kx-x+1$ and $h(x) = x+kx-2kx^2$ in the interval $x \in [0,1]$. Clearly this minimum value will be reached for $f=h$ or for $g=h$, and you can check that the intersection $f=h$ will be higher value than $g=h$, so we only check $g=h$. This means $3kx^2-2kx-2x+1 = 3kx^2-2x(k+1) + 1 = 0$. Discriminant gives us $4(k+1)^2-12k = 4(k^2-k+1)$, so $x = \frac{k+1 \pm \sqrt{k^2-k+1}}{3k}$. In this case $x = \frac{k+1-\sqrt{k^2-k+1}}{3k}$. Plugging this in $g$ will give an equation we can do derivative and we're done. No idea how this would be done nicely.