Answer
$y' = -\dfrac{y^{3/2}}{x^{3/2}}$
Work Step by Step
In order to derivate this function you have to apply implicit differentation method.
First, take the function to it's f(x)=0 form
$x^{-1/2}+y^{-1/2}-1=0$
Then derivate the whole equation. Rember the put y' every time you derivate y
$-\dfrac{1}{2}x^{-3/2}-\dfrac{1}{2}y^{-3/2}y'=0$
*Note: Here you have to apply the product rule twice
Solve for y' and you have the answer
$\dfrac{y'}{-2y^{3/2}} =\dfrac{1}{2x^{3/2}}$
$y' = -\dfrac{y^{3/2}}{x^{3/2}}$
