If $u={\tan }^{-1}\left(\frac{x^3+y^3}{x-y}\right)$,

then $x^2\frac{{\partial }^{2}u}{\partial x^2}+2xy\frac{{\partial }^{2}u}{\partial x\partial y}+y^2\frac{{\partial }^{2}u}{\partial y^2}=?$

The correct option is A $(2\cos 2u-1)\sin 2u$
Given : $u={\tan }^{-1}\left(\frac{x^3+y^3}{x-y}\right)$

$\tan u=\frac{x^3+y^3}{x-y}$

$(x-y)\tan u=x^3+y^3\to \left(1\right)$

Differentiate w.r.t ${}^{\prime }{x}^{\prime }$ partially

$+\tan u+(x-y){\sec }^{2}u\frac{\partial u}{\partial x}=3x^2\to \left(2\right)$

Differntiate w.r.t ${}^{\prime }{y}^{\prime }$ partially

$-\tan u-(x-y){\sec }^{2}u\frac{\partial u}{\partial x}=3x^2\to \left(3\right)$

Further $x\left(2\right)$ + $y\left(3\right)$

$(x-y)\tan \left(u\right)+{\sec }^{2}u(x-y)\{x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\}=3(x^3+y^3)$

$(x-y)[\tan \left(u\right)+{\sec }^{2}u\{x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\}]=3(x^3+y^3)$

$\tan \left(u\right)+{\sec }^{2}u\{x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\}=\frac{3(x^3+y^3)}{(x-y)}$

$\tan \left(u\right)+{\sec }^{2}u\{x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\}=3\tan u$

$\Rightarrow x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\sin \left(2u\right)$

Differentiating w.r.t $x$ and $y$ gives

$\frac{\partial u}{\partial x}+x\frac{{\partial }^{2}u}{\partial x^2}+x\frac{{\partial }^{2}u}{\partial x\partial y}=2\cos \left(2u\right)\frac{\partial u}{\partial x}\to \left(4\right)$

$x\frac{{\partial }^{2}u}{\partial x\partial y}+y\frac{{\partial }^{2}u}{\partial y^2}+\frac{\partial u}{\partial y}=2\cos \left(2u\right)\frac{\partial u}{\partial y}$

Further $x\left(4\right)$ + $y\left(5\right)$ gives

$x^2\frac{{\partial }^{2}u}{\partial x^2}+2xy\frac{{\partial }^{2}u}{\partial x\partial y}+y^2\frac{{\partial }^{2}u}{\partial y^2}=\{2\cos \left(2u\right)-1\}\{x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\}$

$=\{2\cos \left(2u\right)-1\}\sin \left(2u\right)$

Hence option $A$ is correct answer