Yes $ f $ is homogeneous of degree $ n $, $ f (tx, ty) = t ^ {n} f (x, y) $

watch CA

$ f_ {x} (tx, ty) = t ^ {n-1} f_ {x} (x, y) $

My proof is a little misleading as follows:

$ u = tx, v = ty quadric f {x} (tx, ty) = frac { partial f (u, v)} { partial u} cdot frac { partial u} { partial x } + frac { partial f (u, v)} { partial v} cdot frac { partial v} { partial x} = f_ {u} (u, v) cdot t $……. (1)

$ quad quad quad quad quad quad frac { partial} { partial x} (t ^ nf (x, y)) = t ^ nf_ {x} (x, y) $……. (2)

(1) = (2)$ qquad f_ {u} (u, v) = t ^ {n-1} f_ {x} (x, y) $

$ quad quad quad quad , f_ {u} (tx, ty) = t ^ {n-1} f_ {x} (x, y) $

On the last line, the note on the left side is supposed to be $ x $however, I receive $ u $.