The terms used in the following context follow the book Differential Geometry by Do Carmo.
Let $S$ be a regular surface. Let $X(u_1,u_2)=X:U \rightarrow S$ be a parametrization of $S$. Given two vector fields $V_1,V_2: S \rightarrow \mathbb{R}^3$, $p \in S$, the directional derivative of $V_1$ along $V_2$ at $p$ is defined as $D_{V_2}V_1(p)=(V_1 \circ \alpha)'(0)$, where $\alpha:(-\epsilon,\epsilon) \rightarrow S$ is a curve such that $\alpha(0)=p$ and $\alpha'(0)=V_2(p)$. Now, I let $V_1(p)=\frac{\partial X}{\partial u_1}(X^{-1}(p))$ and $V_2(p)=\frac{\partial X}{\partial u_2}(X^{-1}(p))$. I am wondering if it is true that $D_{V_2}V_1=\frac{\partial^2X}{\partial u_2\partial u_1}(X^{-1}(p))$.
I attempted to use the definition, but it fails because I fail to use chain rule for $X$. Do you have any idea how to prove this statement? Or this statement is false?
Let $(u_1^0, u_2^0) = X^{-1}(p)$ and let $\alpha(t) = X(u_1^0, u_2^0+t).$ Then $\alpha(0) = p$ and $\alpha'(0) = \frac{\partial X}{\partial u_2}(u_1^0 u_2^0) = \frac{\partial X}{\partial u_2}(X^{-1}(p)) = V_2(p).$ Thus, $$ D_{V_2} V_1(p) = (V_1 \circ \alpha)'(0) = \left. \frac{d}{dt}\left(\frac{\partial X}{\partial u_1}(X^{-1}(X(u_1^0, u_2^0+t))) \right) \right|_{t=0} = \left. \frac{d}{dt}\left(\frac{\partial X}{\partial u_1}(u_1^0, u_2^0+t) \right) \right|_{t=0} = \left. \frac{\partial}{\partial u_2} \frac{\partial X}{\partial u_1}(u_1^0, u_2^0+t) \right|_{t=0} = \frac{\partial X}{\partial u_2 \partial u_1}(u_1^0, u_2^0) = \frac{\partial X}{\partial u_2 \partial u_1}(X^{-1}(p)) $$