Answer
$x^2\dfrac{\partial ^2 f }{\partial x^2} + 2xy \dfrac{\partial ^2 f }{\partial x \partial y} +y^2 \dfrac{\partial ^2 f }{\partial y^2} =n(n-1) f(x,y)$
Work Step by Step
Since, $\dfrac{\partial f }{\partial x}[\dfrac{d(tx)}{dt}]+\dfrac{\partial f }{\partial y}[\dfrac{d(ty)}{dt}]=nt^{n-1}f(x,y)$
Let us consider $x=\dfrac{d(tx)}{dt}=x$ and $y=\dfrac{d(ty)}{dt}$
This implies that $\dfrac{d^2(tx)}{dt^2}=x$ and $\dfrac{d^2(ty)}{dt^2}=0$
$[x(xf_{xx}+yf_{xy}+(0)f_x]+[y(y(f_{yy}+xf_{xy})+(0)f_y]=n(n-1) t^{n-1} f(x,y)$
or, $[x(xf_{xx}+yf_{xy}]+[y(y(f_{yy}+xf_{xy})]=n(n-1) t^{n-1} f(x,y)$
This gives: $x^2f_{xx} +xyf_{xy} +xy f_{xy}=n \times (n-1) \times t^{n-2} f(x,y)$
Now set $t=1$, then we have
$x^2 \times f_{xx} + 2xy \times f_{xy} +y^2 f_{yy}=n(n-1) f(x,y)$
The result has been proved.
$x^2\dfrac{\partial ^2 f }{\partial x^2} + 2xy \dfrac{\partial ^2 f }{\partial x \partial y} +y^2 \dfrac{\partial ^2 f }{\partial y^2} =n(n-1) f(x,y)$
