Prove that $T_pM^*=\mathfrak m_p/\mathfrak m_p^2$. Let $M$ a manifold of dimension $n$ and $T_pM$ its tangent space. Let denote $T_p^*M=\mathcal L(T_pM,\mathbb R)$ it's dual. Let also denote $$\mathfrak m_p=\{f\in\mathcal C^\infty (M)\mid f(p)=0\}\quad \text{and}\quad \mathfrak m_p^2=\left\{\sum_{i=1}^k f_ig_i\mid f_i,g_i\in\mathfrak m_p\right\}.$$
Show that $T_pM^*=\mathfrak m_p/\mathfrak m_p^2$.
Let $$\omega :\mathfrak m_p\longrightarrow T_p^*M$$
defined by $$\omega (f)=\omega (f)(X)=X(f)$$
for all derivation $X\in T_pM$. It's easy to prove that $\mathfrak m_p^2\subset \ker \omega $. Then I would like to use the first isomorphisme theorem, but I don't know if I can.  
Am I on the right way ? How to continue ?
By the way, what does mean that $T_p^*M=\mathfrak m_p/\mathfrak m_p^2$ ? I can't give a concret representation of it.

Very late but here it is.
Recall that $T_p M$ is the vector space of all derivations at $p$ denoted by $Der C^{\infty}(M)_p$. Now what do these functions look like? Let $\delta\in Der C^{\infty}(M)_p$. First consider the ideal that you introduce: $\mathfrak{m}_p=\{f\in C^{\infty}(M): f(p)=0\}$. This is actually an ideal inside the ring of functions $C^{\infty}(M)_p$. Since this is an ideal we can then square it and consider $\mathfrak{m}_p^2$.
Notice that we have by the Leigniz Rule $\delta(\mathbb{R})=0$ and $\delta(\mathfrak{m}_p^2)=0$. Now this little fact therefore implies the following: there is a map $\Psi: T_p(M) \to (\mathfrak{m}_p \backslash \mathfrak{m}_p^2)^*$ defined by sending a derivation $\delta$ to $(f+\mathfrak{m}_p^2 \mapsto \delta(f))$. $(\mathfrak{m}_p \backslash \mathfrak{m}_p^2)^*$ is simply the dual vector space to $(\mathfrak{m}_p \backslash \mathfrak{m}_p^2)$, i.e the vector space of linear functionals on $(\mathfrak{m}_p \backslash \mathfrak{m}_p^2)$, so the above definition makes sense. The map $\Psi$ is a bijective map! This offers a nice way to intrisincally define the tangent space $T_P(M)$.
From this one can see that $T_P(M)^*$ is the same as $(\mathfrak{m}_p \backslash \mathfrak{m}_p^2)$ and so this what it means. Now Thomas' answer shows the fact.
