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?
> 
> 
> 



Factoring certain Hessians of real homogeneous bivariate polynomials
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,
$$ Q = \frac{n}{n-1} f \Big( \mbox{Hess} f \Big) .$$


