Let $p = (p_1, \ldots, p_n)$ and $x = (x_1, \ldots, x_n)$ be arbitrary vectors such that their standard inner product equals $y$, a constant. (So $p$ depends on $y$ and $x$.) Supose that $x_*, x_*^h$ are functions (with codomain $\mathbb{N}$) such that, for all $p$, we have $$ x_*(p, y) = x_*^h(p, v(p, y)) $$ where $v$ is another function. Show that $$ \frac{\partial x_*^h}{\partial p_k} = \frac{\partial x_*}{\partial p_k} + x_k \frac{\partial x_*}{\partial y} $$ Now first things first: why can this even be true? Must we not leave out the last term? Because by the first centered equality, just take the derivative with respect to $p_k$ and we are done.
Anyway, I have no idea how to solve this. I thought maybe we should take the first centered equality, take the derivative with respect to $p_k$ and write everything out. Then, take the first centered equality again and take the derivative with respect to $y$. And then combine the two equations.
Any help is greatly appreciated.
Just in case you're interested: $x_*$ is the amount of units of $x_1$ you buy (optimizing your satisfaction) given the vector containing the prices of the goods, $p$, and the budget $y$. $x_*^h$ is the amount of units of $x_1$ you buy, minimizing your expenditure, given prices $p$ and satisfaction level $v(p, y)$, the maximum satisfaction you can hope to achieve under the prices and with yiur budget.
It's a little confusing since you don't say what the input arguments to your functions are, so I'll write it out.
First, note that $$ \frac{\partial}{\partial p_k} y = \frac{\partial}{\partial p_k} p^Tx = x_k $$ Then, using your first equality and then the chain rule, we get: $$ \frac{d x_*^h(p,v(p,y))}{d p_k} = \frac{d x_*(p,y)}{d p_k} = \frac{\partial x_*(p,y)}{\partial p_k} + \frac{\partial x_*(p,y)}{\partial y} \frac{\partial y}{\partial p_k} = \frac{\partial x_*(p,y)}{\partial p_k} + \frac{\partial x_*(p,y)}{\partial y} x_k $$ where I've differentiated total derivatives vs partial derivatives.