Answer
$du=\displaystyle \frac{x}{\sqrt{x^{2}+3y^{2}}}dx+\frac{3y}{\sqrt{x^{2}+3y^{2}}}dy$
Work Step by Step
$du=\displaystyle \frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy$
$u=\sqrt{x^{2}+3y^{2}}=(x^{2}+3y^{2})^{1/2}$
$\displaystyle \frac{\partial u}{\partial x}dx=\frac{1}{2}(x^{2}+3y^{2})^{-1/2}(2x)dx=\frac{x}{\sqrt{x^{2}+3y^{2}}}dx$
$\displaystyle \frac{\partial u}{\partial y}dy=\frac{1}{2}(x^{2}+3y^{2})^{-1/2}(6y)dy=\frac{3y}{\sqrt{x^{2}+3y^{2}}}dy$
$du=\displaystyle \frac{x}{\sqrt{x^{2}+3y^{2}}}dx+\frac{3y}{\sqrt{x^{2}+3y^{2}}}dy$