Let consider the following data:
- the family of pairs $(M,p)$ with $M$ a smooth manifold and $p \in M$
- for every pair $(M,p)$ and $(N,q)$ as above a set $\hom[(M,p),(N,q)]$ whose elements are germs of differentiable functions sending $p$ in $q$: spelled out an element of $\hom[(M,p),(N,q)]$ is an equivalence class of mappings $(U,\varphi)$ where $p \in U \subset M$ is an open subset and $\varphi \colon U \to N$ is a differentiable mappings such that $\varphi(p)=q$.
This data define a category where identities are the obvious one and composition is given in the following way: if $[(U,\varphi)] \colon (M,p) \to (N,q)$ and $[(V,\psi)] \colon (N,q) \to (K,x)$ then the composite is given by the class $[\varphi^{-1}(V),\psi \circ \varphi]$. Let call this category $\mathbf {Diff}_*$. Now there's a construction sending every pair $(M,p)$ into $T_p(M)$ the tangent space to $M$ in $p$ and to every mappings $[(U,\varphi)] \colon (M,p) \to (N,q)$ the linear mapping $T([(U,\varphi)]) = d(\varphi)_p \colon T_p(M) \to T_q(N)$ (the differential).
This should be a functor from the category above mentioned to the category of $\mathbf R$-vector spaces.
Here's the question:
Is this functor $$T \colon \mathbf {Diff}_* \to \mathbb{R}\text{-}\mathbf{Vect}$$ an equivalence of categories?
Here's an explicit counterexample (loosely based on the comments): Let $(M,p) = (N,q) = (\mathbb{R}, 0)$. Let $f : (\mathbb{R}, 0) \to (\mathbb{R}, 0)$ be the germ of the function $x \mapsto x^2$. It's more or less clear that this germ is not equal to the germ of the zero function. But $T(f) : T_0\mathbb{R} = \mathbb{R} \to T_0\mathbb{R} = \mathbb{R}$ is the zero function (because the derivative of $x \mapsto x^2$ is zero at $x=0$), and is therefore equal to $T(0)$. This show that your function isn't faithful, and therefore not an equivalence of categories (it's full and essentially surjective, though).