1. rvc

$u=\tan^{-1} \frac{ xy }{ \sqrt{1+x^2+y^2} }$

2. rvc

|dw:1444407988651:dw|

3. IrishBoy123

i'd go with $$u = \tan^{-1} z$$, ie $$z = \frac{ xy }{ \sqrt{1+x^2+y^2} }$$ because $$\dfrac{du }{dz} = \dfrac{1}{1+z^2}$$ and $$\dfrac{∂u}{∂x} = \dfrac{du}{dz}\cdot\dfrac{∂z}{∂x}$$