Eeny, meeny, minimum

$y = x + \sqrt {y}$

$x = y - \sqrt {y} = (y - \sqrt {y} + 0.25) - 0.25 = ( \sqrt {y} - 0.5)^2 - 0.25$

Now, since the minimum value of a square is 0, the minimum value of x is :

$\boxed { -0.25 }$

This minimum value is at:

$\sqrt {y} = 0.5$

$y = 0.25$

We differentiate it with respect to y. We get $1 = \frac{\mathrm d x}{\mathrm d y} + \frac{1}{2*\sqrt{y}}$ . Then, setting $\frac{\mathrm d x}{\mathrm d y}=0$ (since we are searching for the minimum of $x=f(y)=y-\sqrt{y}$ ) we get an answer $y= \frac{1}{4}$ which makes $x=\frac{-1}{4}$