Q. Given that $u=\arctan\left(\frac{x^3+y^3}{x-y}\right)$, prove the following :
$$x^2\frac{\partial^2u}{\partial x^2}+2xy\frac{\partial^2u}{\partial x\partial y}+y^2\frac{\partial^2u}{\partial y^2}=(1-4\sin^2 u)\sin(2u)$$
(The relevant partial derivatives are assumed to be continuous)
Attempted incomplete solution:
$$\tan(u)=\frac{x^3+y^3}{x-y}=f~\textrm{(say)}$$
We note that $f$ is a homogeneous function in $x,y$ of degree $2$ and hence, by a general result of Euler's Theorem, we have,
$$x^2\frac{\partial^2f}{\partial x^2}+2xy\frac{\partial^2f}{\partial x\partial y}+y^2\frac{\partial^2f}{\partial y^2}=2(2-1)f=2\tan(u)$$
I'm having trouble expressing the second order partial derivatives of $f$ in terms of that of $u$ (I'm relatively new at this). Can someone help me out? I don't want the complete solution, just how to apply the chain rule to get the partial derivatives of $f$ in terms of $u$. Thanks.