Writing homogenous function in terms of its second partials. Suppose $f(x,y) \in C^2$ and homogeneous of degree $n$.
I am trying to show $\frac{x^2f_{xx}+2xyf_{xy} + y^2f_{yy}}{n(n-1)} = f$.
I think this result may be a consequence of a multivariate version of the Chain Rule but I am looking for some help to write out a formal derivation to exploit this idea.
 A: You can use this way to show. In fact, let
$$ f(x,y)=\sum_{i=0}^nx^iy^{n-i}.$$
Then
\begin{eqnarray}
&&\frac{x^2f_{xx}+2xyf_{xy} + y^2f_{yy}}{n(n-1)}\\
&=&\frac{1}{n(n-1)}\left[x^2\sum_{i=0}^ni(i-1)x^{i-2}y^{n-i}+2xy\sum_{i=0}^ni(n-i)x^{i-1}y^{n-i-1}+y^2\sum_{i=0}^n(n-i)(n-i-1)x^iy^{n-i-2}\right]\\
&=&\frac{1}{n(n-1)}x^2\sum_{i=0}^n\left[i(i-1)+2i(n-i)+(n-i)(n-i-1)\right]x^iy^{n-i}\\
&=&\sum_{i=0}^nx^iy^{n-i}\\
&=&f(x,y).
\end{eqnarray}
A: Hint: Consider the definition of homogeneity and look at $f(\lambda x, \lambda y)$ as a function of $\lambda$.

Detailed solution in case the hint isn't sufficient.
Definition of homogeneity of degree $n$:
$$
f(\lambda x, \lambda y) = \lambda^n\,f(x,y)
$$
Find first and second partial derivatives with respect to $\lambda$:
$$
\frac{\partial f(\lambda x, \lambda y)}{\partial x}\,x +
\frac{\partial f(\lambda x, \lambda y)}{\partial y}\,y =
n\,\lambda^{n-1}\,f(x,y)
$$
$$
\frac{\partial^2 f(\lambda x, \lambda y)}{\partial x^2}\,x^2 +
2\frac{\partial^2 f(\lambda x, \lambda y)}{\partial x\,\partial y}\,xy +
\frac{\partial^2 f(\lambda x, \lambda y)}{\partial y^2}\,y^2 =
n\,(n-1)\,\lambda^{n-2}\,f(x,y)
$$
Now set $\lambda = 1$:
$$
\frac{\partial^2 f}{\partial x^2}\,x^2 +
2\frac{\partial^2 f}{\partial x\,\partial y}\,xy +
\frac{\partial^2 f}{\partial y^2}\,y^2 =
n\,(n-1)\,f(x,y)
$$
