If \(u=\tan^{-1}\left(\dfrac{x^3+y^3}{x-y}\right)\), prove that \(x^2\dfrac{\partial^2u}{\partial x^2}+2xy\dfrac{\partial^2u}{\partial x\partial y}+y^2\dfrac{\partial^2u}{\partial y^2}=\sin4u-\sin2u=2\cos3u\sin u\).

Answer: With \(g(u)=\sin2u\), the result \(g(u)[g'(u)-1]=\sin2u(2\cos2u-1)=\sin4u-\sin2u=2\cos3u\sin u\) follows.

Let \(f=\tan u=\dfrac{x^3+y^3}{x-y}\), which is homogeneous of degree \(n=3-1=2\).

Step 1 (Euler, first order): \(x f_x+y f_y=2f\Rightarrow\sec^2u\,(x u_x+y u_y)=2\tan u\)

\[x u_x+y u_y=\frac{2\tan u}{\sec^2u}=2\sin u\cos u=\sin2u=g(u).\]

Step 2 (second order corollary): \(x^2u_{xx}+2xy\,u_{xy}+y^2u_{yy}=g(u)\,[g'(u)-1]\).

\(g'(u)=2\cos2u\). Therefore

\[x^2u_{xx}+2xy\,u_{xy}+y^2u_{yy}=\sin2u\,(2\cos2u-1)=2\sin2u\cos2u-\sin2u=\sin4u-\sin2u.\]

Using \(\sin C-\sin D=2\cos\frac{C+D}{2}\sin\frac{C-D}{2}\): \(\sin4u-\sin2u=2\cos3u\sin u\). Hence proved.

(Derivation of the corollary, if asked: differentiate \(x u_x+y u_y=g(u)\) partially w.r.t. \(x\) and \(y\), multiply by \(x\) and \(y\) and add.)