If $u = e^{x+y} + \ln (x^3+y^3-x^2y-xy^2)$, find the value of $$x^2 \dfrac{\partial^2u}{\partial x^2} +2xy \dfrac{\partial^2u}{\partial x \partial y} +y^2 \dfrac{\partial^2u}{\partial y^2} + x \dfrac{\partial u}{\partial x} + y \dfrac{\partial u}{\partial y} $$
Is Euler's theorem applicable here ? $u$ doesn't look homogeneous and working the partials looks like a pain. Is there any smart way ?
If, as commented by Neeraj Bhauryal, you write $$u=e^{x+y} + \ln (x^3+y^3-x^2y-xy^2)=e^{x+y}+\ln (x+y)+2\ln(x-y)$$ and take into account the "symmetry", the problem of partials is quite simple $$u'_x=e^{x+y}+\frac{1}{x+y}+\frac{2}{x-y}$$ $$u'_y=e^{x+y}+\frac{1}{x+y}-\frac{2}{x-y}$$ $$u''_{xx}=e^{x+y}-\frac{1}{(x+y)^2}-\frac{2}{(x-y)^2}$$ $$u''_{yy}=e^{x+y}-\frac{1}{(x+y)^2}-\frac{2}{(x-y)^2}$$ $$u''_{xy}=e^{x+y}-\frac{1}{(x+y)^2}+\frac{2}{(x-y)^2}$$ and, after a few simplifications, the expression just write $$e^{x+y} (x+y) (x+y+1)$$