Помогите решить!!!!!!!!!!!!!

Решите задачу:

$u=sin^2(x-4y)\\\\\dfrac{\partial u}{\partial x}=2\, sin(x-4y)\cdot cos(x-y)=sin(2x-8y)\\\\\dfrac{\partial u}{\partial y}=2\, sin(x-4y)\cdot cos(x-4y)\cdot (-4)=-4\, sin(2x-8y)\\\\\dfrac{\partial ^2u}{\partial x^2}=cos(2x-8y)\cdot 2=2\, cos(2x-8y)\\\\\dfrac{\partial ^2u}{\partial y^2}=-4\, cos(2x-8y)\cdot (-8)=32\, cos(2x-8y)\\\\\\16\cdot \dfrac{\partial ^2u}{\partial x^2}=16\cdot 2\, cos(2x-8y)=32\, cos(2x-8y)=\dfrac{\partial ^2u}{\partial y^2}$