Prove that $df(p)=0$ if p is a local extrema on a differentiable Manifold.

Question

I am supposed to prove that if $p$ is a local extrema of a function $$f: \; M \rightarrow \mathbb{R}$$ where M is a differentiable Manifold that $$df_p = 0$$ Let $x$ be a map around $p$, then $x(p)$ is a local extrema of $$g: \mathbb{R}^n \rightarrow \mathbb{R} =\; f \circ x^{-1}$$ We can identify $dg_y$ with $Dg(y)$ (where D is the total differential). And because g is a function from $\mathbb{R}^n \rightarrow \mathbb{R}$ we know that $$Dg(x(p)) = 0$$ But from the chain rule we also know that: $$dg_y = df_{x^{-1}(y)} \circ dx^{-1}_y$$ $\Rightarrow$ $$ df_{x^{-1}(x(p))} \circ dx^{-1}_{x(p)} = 0$$ Because $x$ is a map, $dx^{-1}$ is a linear isomorphism between the tangent spaces. Especially its non zero and surjective $$\Rightarrow df_{x^{-1}(x(p))} = df_p = 0$$ But I am not really sure if this actually works, i would appreciate some feedback