Answer
$du=\dfrac{x}{\sqrt{x^2+3y^2}} dx + \dfrac{3y}{\sqrt{x^2+3y^2}} dy$
Work Step by Step
Given the function $u=\sqrt{x^2+3y^2}$
The differential form can be evaluated as follows:
$dw=\dfrac{\partial w}{\partial x} dx + \dfrac{\partial w}{\partial y} dy + \dfrac{\partial w}{\partial z} dz$
We need to find the partial derivatives w.r.t. $x$ and $y$ as follows:
$f_x=\dfrac{1}{2}(x^2+3y^2)^{-(1/2)} (2x)=\dfrac{x}{\sqrt{x^2+3y^2}} \\f_y=\dfrac{3y}{\sqrt{x^2+3y^2}}$
Hence, we have
$du=\dfrac{x}{\sqrt{x^2+3y^2}} dx + \dfrac{3y}{\sqrt{x^2+3y^2}} dy$
