Factoring certain Hessians of real homogeneous bivariate polynomials For any homogeneous polynomial $f \in \mathbb R [x,y]$, define the homogeneous polynomial
$$H(f) := \partial_yf^2\partial_x\partial_xf-2\partial_xf\partial_yf\;\partial_x\partial_yf+\partial_xf^2\partial_y\partial_yf$$
which is the Hessian of $f$ applied to the tangent of its level set. I came across it while studying the asymptotics of certain complete open surfaces. 
From numerical tests, it appears that any homogeneous polynomial $f(x,y)$ factors as follows
$$H(f) = G(x,y) \, f$$
for some homogeneous polynomial $G$.

Question: Is there a simple proof of, or a counterexample to, this factorization?

 A: Assume $f(x,y)=x^{n} P(\frac{y}{x})$ for some $n\geq 0$, and $\mathrm{deg}\, P(x)\leq n$. Then 
$$
f_{x}^{2}f_{xx} + f_{y}^{2}f_{yy}-2f_{x}f_{y}f_{xy} = x^{3n-4}nP(t)\left(-P'(t)^{2}(n-1)+n P''(t)P(t) \right)
$$
where $t=y/x$. Then $G(x,y)=x^{2n-4} n\left(-P'(t)^{2}(n-1)+n P''(t)P(t) \right)$ works provided that $n \geq 2$ because one can easily check that if $\mathrm{deg}\,P\leq n$ then   $\mathrm{deg}(-P'(t)^{2}(n-1)+n P''(t)P(t))\leq 2n-4$. Besides notice that the assertion trivially holds for $n=0$ and $n=1$ homogeneous polynomials $f(x,y)$, we can just take $G=0$ 
A: Let $f$ be a homogeneous polynomial of degree $n$.
Define: 
$$Q:= f_{xx} f_y^2 - 2 f_{xy} f_x f_y + f_{yy} f_x^2.$$
Consider $(n-1)^2 Q$ in the following way
$$(n-1)^2 Q = f_{xx} \Big((n-1)^2 f_y^2\Big) - 2 f_{xy} \Big((n-1)f_x\Big) \Big((n-1)f_y\Big) + f_{yy} \Big((n-1)^2 f_x^2\Big)$$
By Euler's Lemma:
$$(n-1) f_y = x f_{xy} + yf_{yy}, \quad
(n-1) f_x = x f_{xx} + yf_{xy}$$
So, replacing these expressions and simplifying the products we obtain
\begin{align*}
(n-1)^2 Q &= -x^2 f_{xx}f_{xy}^2 -y^2 f_{yy}f_{xy}^2 -2 xy f_{xy}^3 +
f_{xx} f_{yy}\Big(y^2 f_{yy} +2xy f_{xy} + x^2 f_{xx} \Big) \\
& = - f_{xy}^2 \Big( x^2 f_{xx} +2xy f_{xy} + y^2 f_{yy} \Big) + n(n-1) f \Big(f_{xx} f_{yy}\Big)\\
& = - n(n-1) f f_{xy}^2 + n(n-1) f (f_{xx} f_{yy})\\
& = n(n-1) f \Big(f_{xx} f_{yy} - f_{xy}^2\Big) \\
& = n(n-1) f \Big( \mbox{Hess} f \Big)
\end{align*}
Finally,
$$ Q = \frac{n}{n-1} f \Big( \mbox{Hess} f \Big) .$$
