Q.Prove that $f(x,y)=x^3-2x^2y+3xy^2+y^3$ is homogeneous; what is the degree? Verify Euler's Theorem for $f$.

Definition (Homogeneous Function).

1. $F:A\to\mathbb R$, $A\subset\mathbb R^2$, is homogeneous of degree $p$ on $A$ if there is a constant $p$ such that

   $F(\lambda x,\lambda y) = \lambda^p F(x,y)$

   for all $\lambda\in\mathbb R$ and suitably restricted $\lambda,x,y$ with $(\lambda x,\lambda y)\in A$.
2. The same definition extends to three variables: $G:B\to\mathbb R$, $B\subset\mathbb R^3$, is homogeneous of degree $p$ if $G(\lambda x,\lambda y,\lambda z)=\lambda^pG(x,y,z)$. ("Suitably restricted" simply means $\lambda,x,y,z$ are kept away from values that would divide by zero.)
   Tip

   The fastest test in practice: a function built purely as a ratio/sum of terms each of the same total degree (every monomial in $x,y,z$ carrying the same power count, arguments of $\sin,\cos,\log,e^{(\cdot)}$ etc. being themselves degree-$0$ ratios) is homogeneous, with degree = (degree of numerator) $-$ (degree of denominator), or the common degree of every term in a sum. A function that mixes terms of genuinely different degrees — including a bare additive constant, which is "degree $0$" — is NOT homogeneous, since $\lambda^p$ cannot simultaneously rescale two differently-scaling pieces correctly for every $\lambda$.

   For instance $F(x,y)=x^3-2y^3+5xy^2$ is homogeneous of degree $3$ (every term has total degree 3), while $G(x,y)=e^{x^2}+3y^2$ is not homogeneous, because $G(\lambda x,\lambda y)=e^{(\lambda x)^2}+3(\lambda y)^2\ne\lambda^pG(x,y)$ for any single $p$ (the exponential term does not scale as a power of $\lambda$ at all). Theorem (Euler, on Homogeneous Functions). If $F:A\to\mathbb R$ has continuous partial derivatives and is homogeneous of degree $p$ on $A\subset\mathbb R^2$, then

   $x\frac{\partial F}{\partial x}(x,y) + y\frac{\partial F}{\partial y}(x,y) = pF(x,y) \qquad \forall\,(x,y)\in A.$

   The three-variable version is exactly analogous: if $F:B\to\mathbb R$ ($B\subset\mathbb R^3$) is homogeneous of degree $p$ with continuous partials,

   $x\frac{\partial F}{\partial x}+y\frac{\partial F}{\partial y}+z\frac{\partial F}{\partial z} = pF(x,y,z) \qquad \forall\,(x,y,z)\in B.$

   (The proof is omitted in this course, but the theorem holds for a homogeneous function of any number of variables, and is a fast route to a first-order-partial-derivative identity without differentiating at all.)
   Tip

   When the given function is not homogeneous but is built as (some function) applied to a homogeneous "core" — e.g. $u=\sin^{-1}\!\left(\dfrac{x+y}{\sqrt x+\sqrt y}\right)$ or $v=\log\!\left(\dfrac{x^2+y^2}{x+y}\right)$ — the working trick is: isolate the homogeneous piece $f$ (here $f=\sin u=\dfrac{x+y}{\sqrt x+\sqrt y}$, degree $\tfrac12$; or $f=e^v=\dfrac{x^2+y^2}{x+y}$, degree $1$), apply Euler's Theorem to that piece ($xf_x+yf_y=pf$), then differentiate the outer function ($\sin u$ or $e^v$) with the chain rule and simplify (dividing by $\cos u$, or by $e^v$) to land on the required identity in $u$ or $v$ directly. This is far shorter than a brute-force computation of every partial derivative of the original (non-homogeneous) function.