Answer
$\dfrac{y}{\sqrt{3}+\sqrt{y}}=\dfrac{y\sqrt{y}-y\sqrt{3}}{y-3}$
Work Step by Step
$\dfrac{y}{\sqrt{3}+\sqrt{y}}$
Multiply both numerator and denominator by $\sqrt{3}-\sqrt{y}$ and simplify:
$\dfrac{y}{\sqrt{3}+\sqrt{y}}=\Big(\dfrac{y}{\sqrt{3}+\sqrt{y}}\Big)\Big(\dfrac{\sqrt{3}-\sqrt{y}}{\sqrt{3}-\sqrt{y}}\Big)=\dfrac{y(\sqrt{3}-\sqrt{y})}{3-y}=...$
$...=\dfrac{y\sqrt{y}-y\sqrt{3}}{y-3}$