Proving that Marshallian demand is of the form: $x_i^*(p,I) = \hat{x}_i^*(p)I$ with certain conditions Can I please have some feedback/help proving the following. My proof is below but I am quite uncertain as to whether my solution is efficient. Thank you.
If $u(x)$ is a homothetic utility, then show that Marshallian demand is of the form: $x_i^*(p,I) = \hat{x}_i^*(p)I.$
$\textit{Proof.}$ If $u(x)$ is homothetic then
$$
\forall \alpha \in \mathbb{R}_{+}, \forall x : \hskip 6pt
\frac{\partial u(x)}{\partial x} = 
\frac{\partial u(\alpha \cdot x)}{\partial x}.
$$
Now suppose $$
x_i^*(p,I) = \hat{x}_i^*(p)I
$$
does not hold, which is equivalent to
$$
u(x_i^*(p,I)) \ne u(\hat{x}_i^*(p)I).
$$
Precisely, $x_i^*(p,I)$ and $\hat{x}_i^*(p)I$ may be set valued. In this case we refer to two elements, with at least one of which is not included in both sets.
Case 1.
$$
u(x_i^*(p,I)) &gt; u(\hat{x}_i^*(p)I)
$$
As $u$ is homothetic
$$
u(x_i^*(p,I)) =  u(I \cdot \frac{1}{I}\cdot x_i^*(p,I)) = I \cdot u(\frac{1}{I}\cdot \hat{x}_i^*(p)I).
$$
Using this we have
$$
I \cdot u(\frac{1}{I}\cdot x_i^*(p,I)) = u(x_i^*(p,I)) &gt; u(I\cdot \hat{x}_i^*(p)I) = I\cdot u(x_i^*(p,I))
$$
thus we have
$$
u(\frac{1}{I}\cdot x_i^*(p,I)) &gt; \hat{x}_i^*(p)I)
$$
However as $\frac{1}{I} \cdot x_i^*(p,I)$ is clearly an element of $B(p)I$ this is impossible as $\hat{x}_i^*(p)I$ gives maximal utility in that budget set.
Case 2.
$$
u(x_i^*(p,I)) &lt; u(\hat{x}_i^*(p)I)
$$
As $\hat{x}_i^*(p)I$ is clearly an element of $B(p,I)$ this is impossible as $x_i^*(p,I)$ gives maximal utility in that budget set.
Thus we have proven that
$$
u(x_i^*(p,I)) = u(\hat{x}_i^*(p)I)
$$
which is equivalent with
$$
x_i^*(p,I) = \hat{x}_i^*(p)I.
$$
You have exactly the right idea. For a clearer and more efficient proof, I would suggest the following:

I am not entirely sure what the index $i$ is referring to. A specific good of the bundle? I don't think you need this.
$\hat x (p) $ denotes an optimal bundle for an income equal to $1$, right? It is clearer to write $\hat x (p, 1)$  in order to be consistent with your notation.
The claim which you want to prove is &quot;If $\hat x (p, 1)$  is optimal at an income of 1, then $I \hat x (p, 1)$ is optimal at an income of $I$&quot; . By the way, it is not generally true that every optimal bundle at $I$ is given by $I \hat x (p, 1)$.
To prove the above claim, you only need to note that $I \hat x (p, 1)$ is feasible at $I$ and then argue that it is weakly preferred to every other $y$ that's feasible at $I$. You basically did this in your answer when you noted that the bundle $y / I$ is feasible at an income of 1, and then used optimality of $\hat x (p, 1)$.
As a final remark: Homotheticity is really a property of preferences, not derivatives of utility representations. Indeed, all you need for your argument is that scaling any two bundles by the same positive scalar does not change the preference between then.