Answer
$f_x (tx,ty)=t^{n-1} f_x(x,y)$
Work Step by Step
We know that the homogeneous function of order $n$ can be defined as follows: $f(tx,ty)=t^n f(x,y)$
Thus, we have $\dfrac{\partial f }{\partial x} (tx,ty) (tx)'_x+\dfrac{\partial f }{\partial y} (tx,ty) (ty)'_x=[t^nf(x,y)]' \times x$
$\implies \dfrac{\partial f }{\partial x} (tx,ty) \times (t)+\dfrac{\partial f }{\partial y} \times (tx,ty) (0)=t^n \times f_x(x,y)$
Therefore, $\dfrac{\partial f }{\partial x} (tx,ty) \times (t)=t^n \times f_x(x,y)$
This gives:
$\dfrac{\partial f }{\partial x} (tx,ty) (t)+\dfrac{\partial f }{\partial y} (tx,ty) (0)=t^nf_x(x,y)$
Thus, we get $f_x (tx,ty)=t^{n-1} f_x(x,y)$ (VERIFIED)
