proposition:
Let $S_1$, $S_2$, $S_3$ be 3-regular surfaces and $f:S_1 \to S_2$ and $g: S_2 \to S_3$ be smooth maps. Then show that $g\circ f :S_1 \to S_3$ be also smooth and $$d_p(g\circ f)=(d_{f(p)}g)\circ (d_pf)$$
My attempts:
Let assume $w\in T_p(S_1)$ be tangent vector at p of a curve $\gamma_1 $on $S_1$.
Then, $\gamma_2= f\circ \gamma_1$ is a curve on $S_2$ with tangent vector $d_pf(w)$ at $f(p)$.
So, $\gamma_3g\circ \gamma_2=g\circ f \circ \gamma_1=(g\circ f) \circ \gamma_1$ Is a curve on $S_3$ with tangent vector $d_{f(p)}g(d_pf(w))$ at $g(f(p)$.
From there, I dont know what I need to do in order to reach the Conclusion. Please show me thanks.