Answer
$f_x (tx,ty)=t^{n-1} f_x(x,y)$
Work Step by Step
The homogeneous function of order $n$ can be expressed as: $f(tx,ty)=t^n f(x,y)$
$\dfrac{\partial f }{\partial x} (tx,ty) (tx)'_x+\dfrac{\partial f }{\partial y} (tx,ty) (ty)'_x=(t^nf(x,y))'x$
and $ (t) \times \dfrac{\partial f }{\partial x} (tx,ty)+\dfrac{\partial f }{\partial y} (tx,ty) (0)=t^n [f_x(x,y)]$
This implies that:
$\dfrac{\partial f }{\partial x} (tx,ty) (t)+\dfrac{\partial f }{\partial y} (tx,ty) (0)=t^nf_x(x,y)$
Hence, it has been verified that $f_x (tx,ty)=t^{n-1} f_x(x,y)$
