Let $M$ be a manifold, $p\in M$ and $(U,\varphi=(x\_1,\dots,x\_n))$, $(V,\psi=(y\_1,\dots,y\_n))$ two local charts such that $p\in U\cap V$ and $x\_1=y\_1$. Is it true that $(\frac{\partial}{\partial x\_1})\_p=(\frac{\partial}{\partial y\_1})\_p$?

My professor says that the answer is yes, and so I build a proof. But then I found the following counterexample:

Let $M=\mathbb{R}^2$, $p=(0,0)$, $U=V=\mathbb{R}^2$, $\varphi(u,v)=(u,v)$ and $\psi(u,v)=(u,u+v)$. Then $x\_1(y\_1,y\_2)=y\_1$ and $x\_2(y\_1,y\_2)=y\_2-y\_1$. Using the formula for the change of basis I've got

$$(\frac{\partial}{\partial y\_1})\_p=\sum\_{i=1}^2(\frac{\partial x\_i}{\partial y\_1})\_p(\frac{\partial}{\partial x\_i})\_p=(\frac{\partial}{\partial x\_2})\_p-(\frac{\partial}{\partial x\_1})\_p\neq (\frac{\partial}{\partial x\_1})\_p$$

What's wrong with this counterexample?