About Characterization of Homothetic Function - Mathematics for Economists by Simon and Blume Chapter 20 Exercise 18(3)

Question

I am studying homothetic function and got the following problem:

Problem

Determine whether the function $x^3y^6 + 3x^2y^4 + 6xy^2 + 9$ is homothetic or not.

Here is my attempt.

My Attempt

I did it in two different ways, but I got two opposite answers. First by using the following definition of a homothetic function, I showed that the above function is homothetic.

Definition$\quad$ A function $v:\mathbb{R}_{+}^{\mathbf{n}} \to \mathbb{R}$ is called homothetic if it is a monotonic transformation of a homogeneous function, that is, if there is a monotonic transformation $z \mapsto g(z)$, where $g:\mathbb{R}_{+} \to \mathbb{R}$, and a homogeneous function $u:\mathbb{R}_{+}^{\mathbf{n}} \to \mathbb{R}_{+}$ such that $v(\mathbf{x}) = g(u(\mathbf{x}))$ for all $\mathbf{x}$ in $\mathbb{R}_{+}^{\mathbf{n}}$.

Answer 1$\quad$ $v(x,y) = x^3y^6 + 3x^2y^4 + 6xy^2 + 9 = (xy^2)^3 + 3(xy^2)^2 + 6xy^2 + 9$ is a monotonic transformation of a homogeneous function. Define $g:\mathbb{R}_{+} \to \mathbb{R}$ by $g(z) = z^3 + 3z^2 + 6z + 9$. Since $g'(z) = 3z^2 + 6z + 6 = 3(z+1)^2 + 3 > 0$, $g$ is strictly increasing. Define $u:\mathbb{R}_{+}^{2} \to \mathbb{R}_{+}$ by $u(x,y) = xy^2$. Then $u$ is homogeneous of degree $3$. Thus, $v = g \circ u$ is homothetic.

However, if I use the following theorem, I would get that $v$ is not homothetic.

Theorem$\quad$ Let $u$ be a $C^1$ function on $\mathbb{R}_{+}^{\mathbf{n}}$. $u$ is homothetic if and only if, for all $\mathbf{x}$ in $\mathbb{R}_{+}^{\mathbf{n}}$, all $t > 0$, and all $i$, $j$, \begin{align*} \frac{\frac{\partial u}{\partial x_i}(t\mathbf{x})}{\frac{\partial u}{\partial x_j}(t\mathbf{x})} = \frac{\frac{\partial u}{\partial x_i}(\mathbf{x})}{\frac{\partial u}{\partial x_j}(\mathbf{x})}. \end{align*}

Answer 2$\quad$ Since for all $x$ in $\mathbb{R}_{+}^{\mathbf{2}}$, \begin{align*} \frac{\frac{\partial v}{\partial x}(tx,ty)}{\frac{\partial v}{\partial y}(tx,ty)} = \frac{3(tx)^2(ty)^6 + 6(tx)(ty)^4 + 6(ty)^2}{6(tx)^3(ty)^5 + 12(tx)^2(ty)^3 + 12(tx)(ty)} \end{align*} which does not equal to \begin{align*} \frac{\frac{\partial v}{\partial x}(x,y)}{\frac{\partial v}{\partial y}(x,y)} = \frac{3x^2y^6 + 6xy^4 + 6y^2}{6x^3y^5 + 12x^2y^3 + 12xy}, \end{align*} for all $t > 0$, we conclude that $v$ is not homothetic.

My Question

Could someone please help me explain why would I get opposite conclusion in the above problem? What am I missing? I really appreciate it!

Answer 1

Proposition 1. For $a>0,b>0,c>0,d>0$, if $\dfrac{a}{b}=\dfrac{c}{d}$ then $\dfrac{a+c}{b+d}=\dfrac{a}{b}$.

Proof is immediate.

Proposition 2. For $a_i>0,b_i>0$, where $i\in\{1,2,\ldots,n\}$, if $\dfrac{a_i}{b_i}=\dfrac{a_1}{b_1}$ (for all $i\in\{1,2,\ldots,n\}$), then $\dfrac{\sum_{i=1}^{n}a_i}{\sum_{i=1}^{n}b_i}=\dfrac{a_1}{b_1}$.

Proof. Use induction on proposition 1.

Using this proposition, it follows that $\dfrac{3(tx)^2(ty)^6}{6(tx)^3(ty)^5} = \dfrac{6(tx)(ty)^4}{12(tx)^2(ty)^3} = \dfrac{6(ty)^2}{12(tx)(ty)}=\dfrac{3x^2y^6}{6x^3y^5}=\dfrac{6xy^4}{12x^2y^3} = \dfrac{6y^2}{12xy}$ implies $\dfrac{3(tx)^2(ty)^6 + 6(tx)(ty)^4 + 6(ty)^2}{6(tx)^3(ty)^5 + 12(tx)^2(ty)^3 + 12(tx)(ty)}=\dfrac{3x^2y^6 + 6xy^4 + 6y^2}{6x^3y^5 + 12x^2y^3 + 12xy}$ for all $t>0$.